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字符串详解

2024-01-10 30
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简介： 字符串详解



字符串类题目是面试一大考点

这类题目算法一般不太复杂，但灵活多变，实现细节较多，容易出错

在实际应用中也比较普遍

属于熟能生巧的一系列题目

无论是Online assessment还是Onsite面试，挂在这类题目上就太可惜了

字符串基础知识

遍历字符串

字符串比较

709.转换成小写字母

//https://leetcode.cn/problems/to-lower-case/
class Solution {
public:
    string toLowerCase(string s) {
        for (char& ch: s) {
            ch = tolower(ch);
        }
        return s;
    }
};
class Solution {
public:
    string toLowerCase(string s) {
        for (char& ch: s) {
            if (ch >= 65 && ch <= 90) {
                ch |= 32;
            }
        }
        return s;
    }
};

58.最后一个单词的长度

//https://leetcode.cn/problems/length-of-last-word/
class Solution {
public:
    int lengthOfLastWord(string s) {
        int index = s.size() - 1;
        while (s[index] == ' ') {
            index--;
        }
        int wordLength = 0;
        while (index >= 0 && s[index] != ' ') {
            wordLength++;
            index--;
        }
        return wordLength;
    }
};

771.宝石与石头

//https://leetcode.cn/problems/jewels-and-stones/
class Solution {
public:
    int numJewelsInStones(string jewels, string stones) {
        int jewelsCount = 0;
        int jewelsLength = jewels.length(), stonesLength = stones.length();
        for (int i = 0; i < stonesLength; i++) {
            char stone = stones[i];
            for (int j = 0; j < jewelsLength; j++) {
                char jewel = jewels[j];
                if (stone == jewel) {
                    jewelsCount++;
                    break;
                }
            }
        }
        return jewelsCount;
    }
};
class Solution {
public:
    int numJewelsInStones(string jewels, string stones) {
        int jewelsCount = 0;
        unordered_set<char> jewelsSet;
        int jewelsLength = jewels.length(), stonesLength = stones.length();
        for (int i = 0; i < jewelsLength; i++) {
            char jewel = jewels[i];
            jewelsSet.insert(jewel);
        }
        for (int i = 0; i < stonesLength; i++) {
            char stone = stones[i];
            if (jewelsSet.count(stone)) {
                jewelsCount++;
            }
        }
        return jewelsCount;
    }
};

387.字符串中的第一个唯一字符

https://leetcode.cn/problems/first-unique-character-in-a-string/
class Solution {
public:
    int firstUniqChar(string s) {
        unordered_map<int, int> frequency;
        for (char ch: s) {
            ++frequency[ch];
        }
        for (int i = 0; i < s.size(); ++i) {
            if (frequency[s[i]] == 1) {
                return i;
            }
        }
        return -1;
    }
};
class Solution {
public:
    int firstUniqChar(string s) {
        unordered_map<int, int> position;
        int n = s.size();
        for (int i = 0; i < n; ++i) {
            if (position.count(s[i])) {
                position[s[i]] = -1;
            }
            else {
                position[s[i]] = i;
            }
        }
        int first = n;
        for (auto [_, pos]: position) {
            if (pos != -1 && pos < first) {
                first = pos;
            }
        }
        if (first == n) {
            first = -1;
        }
        return first;
    }
};
class Solution {
public:
    int firstUniqChar(string s) {
        unordered_map<char, int> position;
        queue<pair<char, int>> q;
        int n = s.size();
        for (int i = 0; i < n; ++i) {
            if (!position.count(s[i])) {
                position[s[i]] = i;
                q.emplace(s[i], i);
            }
            else {
                position[s[i]] = -1;
                while (!q.empty() && position[q.front().first] == -1) {
                    q.pop();
                }
            }
        }
        return q.empty() ? -1 : q.front().second;
    }
};

14.最长公共前缀

//https://leetcode.cn/problems/longest-common-prefix/description/
class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if (!strs.size()) {
            return "";
        }
        string prefix = strs[0];
        int count = strs.size();
        for (int i = 1; i < count; ++i) {
            prefix = longestCommonPrefix(prefix, strs[i]);
            if (!prefix.size()) {
                break;
            }
        }
        return prefix;
    }
    string longestCommonPrefix(const string& str1, const string& str2) {
        int length = min(str1.size(), str2.size());
        int index = 0;
        while (index < length && str1[index] == str2[index]) {
            ++index;
        }
        return str1.substr(0, index);
    }
};
class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if (!strs.size()) {
            return "";
        }
        int length = strs[0].size();
        int count = strs.size();
        for (int i = 0; i < length; ++i) {
            char c = strs[0][i];
            for (int j = 1; j < count; ++j) {
                if (i == strs[j].size() || strs[j][i] != c) {
                    return strs[0].substr(0, i);
                }
            }
        }
        return strs[0];
    }
};
class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if (!strs.size()) {
            return "";
        }
        else {
            return longestCommonPrefix(strs, 0, strs.size() - 1);
        }
    }
    string longestCommonPrefix(const vector<string>& strs, int start, int end) {
        if (start == end) {
            return strs[start];
        }
        else {
            int mid = (start + end) / 2;
            string lcpLeft = longestCommonPrefix(strs, start, mid);
            string lcpRight = longestCommonPrefix(strs, mid + 1, end);
            return commonPrefix(lcpLeft, lcpRight);
        }
    }
    string commonPrefix(const string& lcpLeft, const string& lcpRight) {
        int minLength = min(lcpLeft.size(), lcpRight.size());
        for (int i = 0; i < minLength; ++i) {
            if (lcpLeft[i] != lcpRight[i]) {
                return lcpLeft.substr(0, i);
            }
        }
        return lcpLeft.substr(0, minLength);
    }
};
class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if (!strs.size()) {
            return "";
        }
        int minLength = min_element(strs.begin(), strs.end(), [](const string& s, const string& t) {return s.size() < t.size();})->size();
        int low = 0, high = minLength;
        while (low < high) {
            int mid = (high - low + 1) / 2 + low;
            if (isCommonPrefix(strs, mid)) {
                low = mid;
            }
            else {
                high = mid - 1;
            }
        }
        return strs[0].substr(0, low);
    }
    bool isCommonPrefix(const vector<string>& strs, int length) {
        string str0 = strs[0].substr(0, length);
        int count = strs.size();
        for (int i = 1; i < count; ++i) {
            string str = strs[i];
            for (int j = 0; j < length; ++j) {
                if (str0[j] != str[j]) {
                    return false;
                }
            }
        }
        return true;
    }
};

344.反转字符串

//https://leetcode.cn/problems/reverse-string/
class Solution {
public:
    void reverseString(vector<char>& s) {
        int n = s.size();
        for (int left = 0, right = n - 1; left < right; ++left, --right) {
            swap(s[left], s[right]);
        }
    }
};

541.反转字符串Ⅱ

//https://leetcode.cn/problems/reverse-string-ii/
class Solution {
public:
    string reverseStr(string s, int k) {
        int n = s.length();
        for (int i = 0; i < n; i += 2 * k) {
            reverse(s.begin() + i, s.begin() + min(i + k, n));
        }
        return s;
    }
};

151.反转字符串中的单词

//https://leetcode.cn/problems/reverse-words-in-a-string/
class Solution {
    public String reverseWords(String s) {
        // 除去开头和末尾的空白字符
        s = s.trim();
        // 正则匹配连续的空白字符作为分隔符分割
        List<String> wordList = Arrays.asList(s.split("\\s+"));
        Collections.reverse(wordList);
        return String.join(" ", wordList);
    }
}
class Solution {
public:
    string reverseWords(string s) {
        // 反转整个字符串
        reverse(s.begin(), s.end());
        int n = s.size();
        int idx = 0;
        for (int start = 0; start < n; ++start) {
            if (s[start] != ' ') {
                // 填一个空白字符然后将idx移动到下一个单词的开头位置
                if (idx != 0) s[idx++] = ' ';
                // 循环遍历至单词的末尾
                int end = start;
                while (end < n && s[end] != ' ') s[idx++] = s[end++];
                // 反转整个单词
                reverse(s.begin() + idx - (end - start), s.begin() + idx);
                // 更新start，去找下一个单词
                start = end;
            }
        }
        s.erase(s.begin() + idx, s.end());
        return s;
    }
};
class Solution {
public:
    string reverseWords(string s) {
        int left = 0, right = s.size() - 1;
        // 去掉字符串开头的空白字符
        while (left <= right && s[left] == ' ') ++left;
        // 去掉字符串末尾的空白字符
        while (left <= right && s[right] == ' ') --right;
        deque<string> d;
        string word;
        while (left <= right) {
            char c = s[left];
            if (word.size() && c == ' ') {
                // 将单词 push 到队列的头部
                d.push_front(move(word));
                word = "";
            }
            else if (c != ' ') {
                word += c;
            }
            ++left;
        }
        d.push_front(move(word));
        string ans;
        while (!d.empty()) {
            ans += d.front();
            d.pop_front();
            if (!d.empty()) ans += ' ';
        }
        return ans;
    }
};

557.反转字符串中的单词Ⅲ

//https://leetcode.cn/problems/reverse-words-in-a-string-iii/
class Solution {
public:
    string reverseWords(string s) {
        string ret;
        int length = s.length();
        int i = 0;
        while (i < length) {
            int start = i;
            while (i < length && s[i] != ' ') {
                i++;
            }
            for (int p = start; p < i; p++) {
                ret.push_back(s[start + i - 1 - p]);
            }
            while (i < length && s[i] == ' ') {
                i++;
                ret.push_back(' ');
            }
        }
        return ret;
    }
};
class Solution {
public: 
    string reverseWords(string s) {
        int length = s.length();
        int i = 0;
        while (i < length) {
            int start = i;
            while (i < length && s[i] != ' ') {
                i++;
            }
            int left = start, right = i - 1;
            while (left < right) {
                swap(s[left], s[right]);
                left++;
                right--;
            }
            while (i < length && s[i] == ' ') {
                i++;
            }
        }
        return s;
    }
};

917.仅仅反转字母

//https://leetcode.cn/problems/reverse-only-letters/
class Solution {
public:
    string reverseOnlyLetters(string s) {
        int n = s.size();
        int left = 0, right = n - 1;
        while (true) {
            while (left < right && !isalpha(s[left])) { // 判断左边是否扫描到字母
                left++;
            }
            while (right > left && !isalpha(s[right])) { // 判断右边是否扫描到字母
                right--;
            }
            if (left >= right) {
                break;
            }
            swap(s[left], s[right]);
            left++;
            right--;
        }
        return s;
    }
};

8.字符串转换整数(atoi)

//https://leetcode.cn/problems/string-to-integer-atoi/
class Solution {
public:
    int myAtoi(string s) {
        int index = 0;
        while( index < s.size() && isspace(s[index])) index++;
        int sign = 1;
        if( index < s.size() && s[index] == '-' || s[index] == '+') {
            sign = s[index] == '-' ? -1 : 1;
            index++;
        }
        long long val = 0;
        while(index < s.size() && isdigit(s[index])) {
            if( val  > (2147483647 - (s[index] - '0')) / 10) {
                if (sign == 1) return 2147483647;
                else return -2147483648;
            }
            val = val * 10 + s[index] - '0';
            index++;
        }
        return sign * val;
    }
};
class Automaton {
    string state = "start";
    unordered_map<string, vector<string>> table = {
        {"start", {"start", "signed", "in_number", "end"}},
        {"signed", {"end", "end", "in_number", "end"}},
        {"in_number", {"end", "end", "in_number", "end"}},
        {"end", {"end", "end", "end", "end"}}
    };
    int get_col(char c) {
        if (isspace(c)) return 0;
        if (c == '+' or c == '-') return 1;
        if (isdigit(c)) return 2;
        return 3;
    }
public:
    int sign = 1;
    long long ans = 0;
    void get(char c) {
        state = table[state][get_col(c)];
        if (state == "in_number") {
            ans = ans * 10 + c - '0';
            ans = sign == 1 ? min(ans, (long long)INT_MAX) : min(ans, -(long long)INT_MIN);
        }
        else if (state == "signed")
            sign = c == '+' ? 1 : -1;
    }
};
class Solution {
public:
    int myAtoi(string str) {
        Automaton automaton;
        for (char c : str)
            automaton.get(c);
        return automaton.sign * automaton.ans;
    }
};

Rabin-Karp字符串哈希算法

https://blog.csdn.net/qq_46118239/article/details/126594215

典型的字符串处理:子串、回文、同构

125.验证回文串

//https://leetcode.cn/problems/valid-palindrome/
class Solution {
public:
    bool isPalindrome(string s) {
        string sgood;
        for (char ch: s) {
            if (isalnum(ch)) {
                sgood += tolower(ch);
            }
        }
        string sgood_rev(sgood.rbegin(), sgood.rend());
        return sgood == sgood_rev;
    }
};
class Solution {
public:
    bool isPalindrome(string s) {
        string sgood;
        for (char ch: s) {
            if (isalnum(ch)) {
                sgood += tolower(ch);
            }
        }
        int n = sgood.size();
        int left = 0, right = n - 1;
        while (left < right) {
           if (sgood[left] != sgood[right]) {
                return false;
            }
            ++left;
            --right;
        }
        return true;
    }
};
class Solution {
public:
    bool isPalindrome(string s) {
        int n = s.size();
        int left = 0, right = n - 1;
        while (left < right) {
            while (left < right && !isalnum(s[left])) {
                ++left;
            }
            while (left < right && !isalnum(s[right])) {
                --right;
            }
            if (left < right) {
                if (tolower(s[left]) != tolower(s[right])) {
                    return false;
                }
                ++left;
                --right;
            }
        }
        return true;
    }
};

680.验证回文串Ⅱ

//https://leetcode.cn/problems/valid-palindrome-ii/
class Solution {
public:
    bool checkPalindrome(const string& s, int low, int high) {
        for (int i = low, j = high; i < j; ++i, --j) {
            if (s[i] != s[j]) {
                return false;
            }
        }
        return true;
    }
    bool validPalindrome(string s) {
        int low = 0, high = s.size() - 1;
        while (low < high) {
            char c1 = s[low], c2 = s[high];
            if (c1 == c2) {
                ++low;
                --high;
            } else {
                return checkPalindrome(s, low, high - 1) || checkPalindrome(s, low + 1, high);
            }
        }
        return true;
    }
};

5.最长回文子串

#include <iostream>
#include <string>
#include <vector>
using namespace std;
class Solution {
//https://leetcode.cn/problems/longest-palindromic-substring/
public:
    string longestPalindrome(string s) {
        int n = s.size();
        if (n < 2) {
            return s;
        }
        int maxLen = 1;
        int begin = 0;
        // dp[i][j] 表示 s[i..j] 是否是回文串
        vector<vector<int>> dp(n, vector<int>(n));
        // 初始化：所有长度为 1 的子串都是回文串
        for (int i = 0; i < n; i++) {
            dp[i][i] = true;
        }
        // 递推开始
        // 先枚举子串长度
        for (int L = 2; L <= n; L++) {
            // 枚举左边界，左边界的上限设置可以宽松一些
            for (int i = 0; i < n; i++) {
                // 由 L 和 i 可以确定右边界，即 j - i + 1 = L 得
                int j = L + i - 1;
                // 如果右边界越界，就可以退出当前循环
                if (j >= n) {
                    break;
                }
                if (s[i] != s[j]) {
                    dp[i][j] = false;
                } else {
                    if (j - i < 3) {
                        dp[i][j] = true;
                    } else {
                        dp[i][j] = dp[i + 1][j - 1];
                    }
                }
                // 只要 dp[i][L] == true 成立，就表示子串 s[i..L] 是回文，此时记录回文长度和起始位置
                if (dp[i][j] && j - i + 1 > maxLen) {
                    maxLen = j - i + 1;
                    begin = i;
                }
            }
        }
        return s.substr(begin, maxLen);
    }
};
class Solution {
public:
    pair<int, int> expandAroundCenter(const string& s, int left, int right) {
        while (left >= 0 && right < s.size() && s[left] == s[right]) {
            --left;
            ++right;
        }
        return {left + 1, right - 1};
    }
    string longestPalindrome(string s) {
        int start = 0, end = 0;
        for (int i = 0; i < s.size(); ++i) {
            auto [left1, right1] = expandAroundCenter(s, i, i);
            auto [left2, right2] = expandAroundCenter(s, i, i + 1);
            if (right1 - left1 > end - start) {
                start = left1;
                end = right1;
            }
            if (right2 - left2 > end - start) {
                start = left2;
                end = right2;
            }
        }
        return s.substr(start, end - start + 1);
    }
};
class Solution {
public:
    int expand(const string& s, int left, int right) {
        while (left >= 0 && right < s.size() && s[left] == s[right]) {
            --left;
            ++right;
        }
        return (right - left - 2) / 2;
    }
    string longestPalindrome(string s) {
        int start = 0, end = -1;
        string t = "#";
        for (char c: s) {
            t += c;
            t += '#';
        }
        t += '#';
        s = t;
        vector<int> arm_len;
        int right = -1, j = -1;
        for (int i = 0; i < s.size(); ++i) {
            int cur_arm_len;
            if (right >= i) {
                int i_sym = j * 2 - i;
                int min_arm_len = min(arm_len[i_sym], right - i);
                cur_arm_len = expand(s, i - min_arm_len, i + min_arm_len);
            } else {
                cur_arm_len = expand(s, i, i);
            }
            arm_len.push_back(cur_arm_len);
            if (i + cur_arm_len > right) {
                j = i;
                right = i + cur_arm_len;
            }
            if (cur_arm_len * 2 + 1 > end - start) {
                start = i - cur_arm_len;
                end = i + cur_arm_len;
            }
        }
        string ans;
        for (int i = start; i <= end; ++i) {
            if (s[i] != '#') {
                ans += s[i];
            }
        }
        return ans;
    }
};

205.同构字符串

//https://leetcode.cn/problems/isomorphic-strings/
class Solution {
public:
    bool isIsomorphic(string s, string t) {
        unordered_map<char, char> s2t;
        unordered_map<char, char> t2s;
        int len = s.length();
        for (int i = 0; i < len; ++i) {
            char x = s[i], y = t[i];
            if ((s2t.count(x) && s2t[x] != y) || (t2s.count(y) && t2s[y] != x)) {
                return false;
            }
            s2t[x] = y;
            t2s[y] = x;
        }
        return true;
    }
};

242.有效的字母异位词

//https://leetcode.cn/problems/valid-anagram/
class Solution {
public:
    bool isAnagram(string s, string t) {
        if (s.length() != t.length()) {
            return false;
        }
        sort(s.begin(), s.end());
        sort(t.begin(), t.end());
        return s == t;
    }
};
class Solution {
public:
    bool isAnagram(string s, string t) {
        if (s.length() != t.length()) {
            return false;
        }
        vector<int> table(26, 0);
        for (auto& ch: s) {
            table[ch - 'a']++;
        }
        for (auto& ch: t) {
            table[ch - 'a']--;
            if (table[ch - 'a'] < 0) {
                return false;
            }
        }
        return true;
    }
};

49.字母异位词分组

//https://leetcode.cn/problems/group-anagrams/
class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        for(string& s : strs) {
            string copy = s;
            sort(copy.begin(), copy.end());
            groups[copy].push_back(s);
        }
        vector<vector<string>> ans;
        for(const pair<string, vector<string>> & group : groups) {
            ans.push_back(group.second);
        }
        return ans;
    }
private:
    unordered_map<string, vector<string>> groups;
};
class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        unordered_map<string, vector<string>> mp;
        for (string& str: strs) {
            string key = str;
            sort(key.begin(), key.end());
            mp[key].emplace_back(str);
        }
        vector<vector<string>> ans;
        for (auto it = mp.begin(); it != mp.end(); ++it) {
            ans.emplace_back(it->second);
        }
        return ans;
    }
};
class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        // 自定义对 array<int, 26> 类型的哈希函数
        auto arrayHash = [fn = hash<int>{}] (const array<int, 26>& arr) -> size_t {
            return accumulate(arr.begin(), arr.end(), 0u, [&](size_t acc, int num) {
                return (acc << 1) ^ fn(num);
            });
        };
        unordered_map<array<int, 26>, vector<string>, decltype(arrayHash)> mp(0, arrayHash);
        for (string& str: strs) {
            array<int, 26> counts{};
            int length = str.length();
            for (int i = 0; i < length; ++i) {
                counts[str[i] - 'a'] ++;
            }
            mp[counts].emplace_back(str);
        }
        vector<vector<string>> ans;
        for (auto it = mp.begin(); it != mp.end(); ++it) {
            ans.emplace_back(it->second);
        }
        return ans;
    }
};

438.找到字符串中所有字母异位词

//https://leetcode.cn/problems/find-all-anagrams-in-a-string/
class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        int sLen = s.size(), pLen = p.size();
        if (sLen < pLen) {
            return vector<int>();
        }
        vector<int> ans;
        vector<int> sCount(26);
        vector<int> pCount(26);
        for (int i = 0; i < pLen; ++i) {
            ++sCount[s[i] - 'a'];
            ++pCount[p[i] - 'a'];
        }
        if (sCount == pCount) {
            ans.emplace_back(0);
        }
        for (int i = 0; i < sLen - pLen; ++i) {
            --sCount[s[i] - 'a'];
            ++sCount[s[i + pLen] - 'a'];
            if (sCount == pCount) {
                ans.emplace_back(i + 1);
            }
        }
        return ans;
    }
};
class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        int sLen = s.size(), pLen = p.size();
        if (sLen < pLen) {
            return vector<int>();
        }
        vector<int> ans;
        vector<int> count(26);
        for (int i = 0; i < pLen; ++i) {
            ++count[s[i] - 'a'];
            --count[p[i] - 'a'];
        }
        int differ = 0;
        for (int j = 0; j < 26; ++j) {
            if (count[j] != 0) {
                ++differ;
            }
        }
        if (differ == 0) {
            ans.emplace_back(0);
        }
        for (int i = 0; i < sLen - pLen; ++i) {
            if (count[s[i] - 'a'] == 1) {  // 窗口中字母 s[i] 的数量与字符串 p 中的数量从不同变得相同
                --differ;
            } else if (count[s[i] - 'a'] == 0) {  // 窗口中字母 s[i] 的数量与字符串 p 中的数量从相同变得不同
                ++differ;
            }
            --count[s[i] - 'a'];
            if (count[s[i + pLen] - 'a'] == -1) {  // 窗口中字母 s[i+pLen] 的数量与字符串 p 中的数量从不同变得相同
                --differ;
            } else if (count[s[i + pLen] - 'a'] == 0) {  // 窗口中字母 s[i+pLen] 的数量与字符串 p 中的数量从相同变得不同
                ++differ;
            }
            ++count[s[i + pLen] - 'a'];
            if (differ == 0) {
                ans.emplace_back(i + 1);
            }
        }
        return ans;
    }
};

10.i正则表达式匹配

//https://leetcode.cn/problems/regular-expression-matching/
class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.size();
        int n = p.size();
        auto matches = [&](int i, int j) {
            if (i == 0) {
                return false;
            }
            if (p[j - 1] == '.') {
                return true;
            }
            return s[i - 1] == p[j - 1];
        };
        vector<vector<int>> f(m + 1, vector<int>(n + 1));
        f[0][0] = true;
        for (int i = 0; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (p[j - 1] == '*') {
                    f[i][j] |= f[i][j - 2];
                    if (matches(i, j - 1)) {
                        f[i][j] |= f[i - 1][j];
                    }
                }
                else {
                    if (matches(i, j)) {
                        f[i][j] |= f[i - 1][j - 1];
                    }
                }
            }
        }
        return f[m][n];
    }
};

44.通配符匹配

//https://leetcode.cn/problems/wildcard-matching/
class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.size();
        int n = p.size();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1));
        dp[0][0] = true;
        for (int i = 1; i <= n; ++i) {
            if (p[i - 1] == '*') {
                dp[0][i] = true;
            }
            else {
                break;
            }
        }
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (p[j - 1] == '*') {
                    dp[i][j] = dp[i][j - 1] | dp[i - 1][j];
                }
                else if (p[j - 1] == '?' || s[i - 1] == p[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1];
                }
            }
        }
        return dp[m][n];
    }
};
class Solution {
public:
    bool isMatch(string s, string p) {
        auto allStars = [](const string& str, int left, int right) {
            for (int i = left; i < right; ++i) {
                if (str[i] != '*') {
                    return false;
                }
            }
            return true;
        };
        auto charMatch = [](char u, char v) {
            return u == v || v == '?';
        };
        while (s.size() && p.size() && p.back() != '*') {
            if (charMatch(s.back(), p.back())) {
                s.pop_back();
                p.pop_back();
            }
            else {
                return false;
            }
        }
        if (p.empty()) {
            return s.empty();
        }
        int sIndex = 0, pIndex = 0;
        int sRecord = -1, pRecord = -1;
        while (sIndex < s.size() && pIndex < p.size()) {
            if (p[pIndex] == '*') {
                ++pIndex;
                sRecord = sIndex;
                pRecord = pIndex;
            }
            else if (charMatch(s[sIndex], p[pIndex])) {
                ++sIndex;
                ++pIndex;
            }
            else if (sRecord != -1 && sRecord + 1 < s.size()) {
                ++sRecord;
                sIndex = sRecord;
                pIndex = pRecord;
            }
            else {
                return false;
            }
        }
        return allStars(p, pIndex, p.size());
    }
};

115.不同的子序列

//https://leetcode.cn/problems/distinct-subsequences/
class Solution {
public:
    int numDistinct(string s, string t) {
        int m = s.length(), n = t.length();
        if (m < n) {
            return 0;
        }
        vector<vector<unsigned long long>> dp(m + 1, vector<unsigned long long>(n + 1));
        for (int i = 0; i <= m; i++) {
            dp[i][n] = 1;
        }
        for (int i = m - 1; i >= 0; i--) {
            char sChar = s.at(i);
            for (int j = n - 1; j >= 0; j--) {
                char tChar = t.at(j);
                if (sChar == tChar) {
                    dp[i][j] = dp[i + 1][j + 1] + dp[i + 1][j];
                } else {
                    dp[i][j] = dp[i + 1][j];
                }
            }
        }
        return dp[0][0];
    }
};

字符串模式匹配

字符串匹配

Rabin-Karp 字符串匹配

KMP模式匹配算法

时间复杂度?

看似是两重循环

在上面代码的while循环中，j的值不断减小

减小次数≤每层for循环开始时j的值– while循环结束时j的值

而在每层for循环中，j的值至多增加1

因为j始终非负，所以在整个计算过程中，j减小的幅度总和不会超过j增加的幅度总和

故j的总变化次数至多为2(N＋M)

整个算法的时间复杂度为O(N＋M)

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