LeetCode：5_Longest Palindromic Substring | 最长的回文子串 | Medium

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.

 1 //动态规划法
 2 string LongestPalindromicStringDP(string str)
 3 {
 4     size_t n = str.size();
 5     if (n == 0 || n == 1)
 6         return str;
 7     //创建dp二维数组
 8     //d[i][j] = dp[i+1][j-1] && s[i] == s[j]
 9     bool **dp = new bool *[n];
10     for (size_t i = 0; i < n; ++ i)
11         dp[i] = new bool[n];
12     
13     int nLongestIndex = 0; //最长回文子串的开始下标
14     int nMaxLen = 0; //长度
15 
16     for (size_t i = 0; i < n; i ++)
17         for (size_t j = 0; j < n; j ++)
18             dp[i][j] = false;
19     for (size_t i = 0; i < n; ++ i) 
20         dp[i][i] = true; //初始化一个字母
21 
22     for (size_t i = 0; i < n - 1; ++ i) {
23         if (str[i] == str[i+1]) {
24             dp[i][i+1] = true; //初始化两个相同的字母"aa"
25             nLongestIndex = i;
26             nMaxLen = 2;
27         }
28     }
29     //从长度3开始操作, (aba)ba, a(bab)a, ab(aba)
30     for (size_t len = 3; len <= n; ++ len) {
31         for (size_t i = 0; i < n-len+1; ++ i) {
32             size_t j = i + len - 1;
33             if (str[i] == str[j] && dp[i+1][j-1]){
34                 dp[i][j] = true;
35                 nLongestIndex = i;
36                 nMaxLen = len;
37             }
38         }
39     }
40 
41     //释放dp
42     for (size_t i = 0; i < n; ++ i)
43         delete []dp[i];
44     delete []dp;
45     
46     return str.substr(nLongestIndex, nMaxLen);
47 }

 1 //中心扩散法
 2 string LongestPalindromicString(string str) 
 3 {
 4     size_t n = str.size();
 5     if (n == 0 || n == 1) 
 6         return str;
 7 
 8     size_t start = 0;
 9     size_t nMaxLen = 0;
10 
11     for (size_t i = 0; i < n; ++ i) {
12         size_t j = i, k = i; //分别从中心向两边扩散
13         while(k < n-1 && str[k] == str[k+1])
14             k++; //有相同字母的情况："aaa"
15         while(j > 0 && k < n-1 && str[j-1] == str[k+1]){
16             k++; //不同字母情况: "aba"
17             j--;
18         }
19         if(k-j+1 > nMaxLen){
20             nMaxLen = k-j+1;
21             start = j;
22         }
23     }
24     return str.substr(start, nMaxLen);
25 }