【LeetCode】005. Longest Palindromic Substring

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example:

Input: "babad"

Output: "bab"

Note: "aba" is also a valid answer.

Example:

Input: "cbbd"

Output: "bb"

题解：

Solution 1

　　暴力搜索，所有可能，注意到"bab"和"baab"两种类型的回文字符串即可。

 class Solution {
 public:
     string longestPalindrome(string s) {
         int start = , len = ;
         int n = s.size();
         if (n < )
             return s;
         for (int i = ; i < n - ; ++i) {
             rangeOfPalindrome(s, i, i + , start, len); // "baab"
             rangeOfPalindrome(s, i, i, start, len);   // "bab"
         }
         return s.substr(start, len);
     }

     void rangeOfPalindrome(const string& s, int left, int right, int& start, int& len) {
         int length = len;
         int step = ;
         while ((left - step >= ) && (right + step < s.size())) {
             if (s[left - step] != s[right + step])
                 break;
             ++step;
         }
         length =  * (step - ) + right - left + ;
         if (length > len) {
             len = length;
             start = left - (step - );
         }
     }

 };

　　代码简化：

 class Solution {
 public:
     string longestPalindrome(string s) {
         string res;
         int n = s.size(), idx = , len = ;
         for (int i = ; i < n; ++i) {
             search(s, i, i, idx, len);
             search(s, i, i + , idx, len);
         }
         return s.substr(idx, len);
     }
     void search(const string& s, int i, int j, int& idx, int& len) {
         while (i >=  && j < s.size() && s[i] == s[j]) {
             --i;
             ++j;
         }
         if (len < j - i - ) {
             len = j - i - ;
             idx = i + ;
         }
     }
 };

Soluion 2

　　DP问题。一个长度为 n（n>1) 的回文字符串S（s1, s2,...,sn)，若将字符s0和sn+1分别放置在S的首尾，此时如果s0 == sn+1,那么新的字符串S'也一定是回文字符串。

　　那么递归式为 dp[i][j] = 1                                            if i == j

　　　　　　　　　　　= s[i] == s[j]　                           if i + 1 = j

= s[i] == s[j] && dp[i + 1][j - 1]   if i + 1 < j

 class Solution {
 public:
     string longestPalindrome(string s) {
         int n = s.size();
         if (n < )
             return s;
         int len = , start = ;
         vector<vector<int>> dp(n, vector<int>(n, ));

         for (int i = ; i < n; ++i) {
             for (int j = ; j <= i; ++j) {
                 dp[j][i] = (s[i] == s[j]) && (i - j <=  || dp[j + ][i - ]);
                 if (dp[j][i] && len < i - j + ) {
                     len = i - j + ;
                     start = j;
                 }
             }
         }

         return s.substr(start, len);
     }
 };

　　Manacher算法

Solution 3