1. 全排列
给定一个 没有重复 数字的序列，返回其所有可能的全排列。
示例:
输入: [1,2,3]
输出: [[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1]]
以下程序实现了这一功能，请你填补空白处内容：
···
#include <bits/stdc++.h> using namespace std; class Solution { public: vector<vector<int>> permute(vector<int> &nums) { vector<vector<int>> res; vector<bool> used(nums.size()); dfs(nums, used, res); return res; } private: vector<int> stack; void dfs(vector<int> &nums, vector<bool> &used, vector<vector<int>> &res) { if (stack.size() == nums.size()) { res.push_back(stack); } else { for (int i = 0; i < nums.size(); i++) { if (!used[i]) { _______________; } } } } }; ···
出处：
https://edu.csdn.net/practice/27810770
代码：
#include <bits/stdc++.h> using namespace std; class Solution { public: vector<vector<int>> permute(vector<int> &nums) { vector<vector<int>> res; vector<bool> used(nums.size()); dfs(nums, used, res); return res; } private: vector<int> stack; void dfs(vector<int> &nums, vector<bool> &used, vector<vector<int>> &res) { if (stack.size() == nums.size()) { res.push_back(stack); } else { for (int i = 0; i < nums.size(); i++) { if (!used[i]) { used[i] = true; stack.push_back(nums[i]); dfs(nums, used, res); stack.pop_back(); used[i] = false; } } } } }; string Vector2dToString(vector<vector<int>> vec2d, string sep = ",") { stringstream ss; ss << "["; for (int i = 0; i < vec2d.size(); ++i) { ss << "["; copy(vec2d[i].begin(), vec2d[i].end(), ostream_iterator<int>(ss, sep.c_str())); ss.seekp(-(int)sep.size(), ios_base::end); ss << "]" << sep; } ss.seekp(-(int)sep.size(), ios_base::end); ss << "]"; return ss.str(); } int main() { Solution s; vector<int> nums = {1,2,3}; cout << Vector2dToString(s.permute(nums)) << endl; return 0; }
输出：
[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
2. 分数到小数
给定两个整数，分别表示分数的分子 numerator 和分母 denominator，以 字符串形式返回小数 。
如果小数部分为循环小数，则将循环的部分括在括号内。
如果存在多个答案，只需返回 任意一个 。
对于所有给定的输入，保证 答案字符串的长度小于 104 。
示例 1：
输入：numerator = 1, denominator = 2
输出："0.5"
示例 2：
输入：numerator = 2, denominator = 1
输出："2"
示例 3：
输入：numerator = 2, denominator = 3
输出："0.(6)"
示例 4：
输入：numerator = 4, denominator = 333
输出："0.(012)"
示例 5：
输入：numerator = 1, denominator = 5
输出："0.2"
提示：
-2^31 <= numerator, denominator <= 2^31 - 1
denominator != 0
出处：
https://edu.csdn.net/practice/27810771
代码：
#include <bits/stdc++.h> using namespace std; class Solution { public: string fractionToDecimal(int numerator, int denominator) { if (denominator == 0) return ""; string ans; if ((numerator > 0 && denominator < 0) || (numerator < 0 && denominator > 0)) ans = "-"; long long num = abs(static_cast<long long>(numerator)); long long den = abs(static_cast<long long>(denominator)); long long quo = num / den; long long rem = num % den; ans = ans + to_string(quo); if (!rem) return ans; ans += "."; unordered_map<long long, int> u_m; string s = ""; int pos = 0; while (rem) { if (u_m.find(rem) != u_m.end()) { s.insert(u_m[rem], "("); s += ")"; return ans + s; } u_m[rem] = pos++; s += to_string((rem * 10) / den); rem = (rem * 10) % den; } return ans + s; } }; int main() { Solution s; cout << s.fractionToDecimal(1, 2) << endl; cout << s.fractionToDecimal(2, 1) << endl; cout << s.fractionToDecimal(2, 3) << endl; cout << s.fractionToDecimal(4, 333) << endl; cout << s.fractionToDecimal(1, 5) << endl; return 0; }
输出：
0.5
2
0.(6)
0.(012)
0.2
3. 删除排序链表中的重复元素 II （C语言）
存在一个按升序排列的链表，给你这个链表的头节点 head ，请你删除链表中所有存在数字重复情况的节点，只保留原始链表中 没有重复出现 的数字。
返回同样按升序排列的结果链表。
示例 1：
输入：head = [1,2,3,3,4,4,5]
输出：[1,2,5]
示例 2：
输入：head = [1,1,1,2,3]
输出：[2,3]
提示：
链表中节点数目在范围 [0, 300] 内
-100 <= Node.val <= 100
题目数据保证链表已经按升序排列
以下程序实现了这一功能，请你填补空白处内容：
structListNode{ intval; structListNode*next; }; structListNode*deleteDuplicates(structListNode*head) { structListNodedummy; structListNode*p, *q, *prev; prev=&dummy; dummy.next=head; p=q=head; while (p!=NULL) { _____________________; } returndummy.next; } intmain(intargc, char**argv) { inti; structListNode*head=NULL; structListNode*prev=NULL; structListNode*p; for (i=0; i<argc-1; i++) { p=malloc(sizeof(*p)); p->val=atoi(argv[i+1]); p->next=NULL; if (head==NULL) { head=p; prev=head; } else { prev->next=p; prev=p; } } p=deleteDuplicates(head); while (p!=NULL) { printf("%d ", p->val); p=p->next; } printf("\n"); return0; } ```
出处：
https://edu.csdn.net/practice/27810772
代码： (C语言)
structListNode{ intval; structListNode*next; }; structListNode*deleteDuplicates(structListNode*head) { structListNodedummy; structListNode*p, *q, *prev; prev=&dummy; dummy.next=head; p=q=head; while (p!=NULL) { while (q!=NULL&&q->val==p->val) { q=q->next; } if (p->next==q) { prev=p; } else { prev->next=q; } p=q; } returndummy.next; } intmain() { inti; structListNode*head=NULL; structListNode*prev=NULL; structListNode*p; intnums[] = {1,2,3,3,4,4,5}; intcount=sizeof(nums) /sizeof(nums[0]); for (i=0; i<count; i++) { p= (ListNode*)malloc(sizeof(*p)); p->val=nums[i]; p->next=NULL; if (head==NULL) { head=p; prev=head; } else { prev->next=p; prev=p; } } p=deleteDuplicates(head); while (p!=NULL) { printf("%d->", p->val); p=p->next; } printf("null\n"); return0; }
输出：
1->2->5->null
单链表的创建、打印写成函数形式：
#include <stdio.h> #include <stdlib.h> #define Test(nums) {\ count = sizeof(nums) / sizeof(nums[0]);\ head = arrayToList(nums, count);\ printList(head);\ head = deleteDuplicates(head);\ printList(head);} struct ListNode { int val; struct ListNode *next; }; struct ListNode *deleteDuplicates(struct ListNode *head) { struct ListNode dummy; struct ListNode *p, *q, *prev; prev = &dummy; dummy.next = head; p = q = head; while (p != NULL) { while (q != NULL && q->val == p->val) { q = q->next; } if (p->next == q) { prev = p; } else { prev->next = q; } p = q; } return dummy.next; } struct ListNode *arrayToList(int* nums, int count) { struct ListNode *head = NULL; struct ListNode *prev = NULL; struct ListNode *p; if (count == 0) return NULL; for (int i = 0; i < count; i++) { p = (ListNode*)malloc(sizeof(*p)); p->val = nums[i]; p->next = NULL; if (head == NULL) { head = p; prev = head; } else { prev->next = p; prev = p; } } return head; } void printList(ListNode *head) { ListNode *p = head; while (p != NULL) { printf("%d->", p->val); p = p->next; } printf("null\n"); } int main() { struct ListNode *head; int count; int head1[] = {1,2,3,3,4,4,5}; Test(head1) int head2[] = {1,1,1,2,3}; Test(head2) return 0; }
输出：
1->2->3->3->4->4->5->null
1->2->5->null
1->1->1->2->3->null
2->3->null
