string+DP leetcode-4.最长回文子串

5. Longest Palindromic Substring

题面

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

找出给定数组中，最长的回文子串。最长输入尾1000。

样例

Example 1:

Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.

Example 2:

Input: "cbbd"
Output: "bb"

Example 3:

Input: "abcd"
Output: "a"

Example 4:

Input: "eeeeeeeeeeeeeeeeeeeeee"
Output: "eeeeeeeeeeeeeeeeeeeeee"

思路1：暴力，不可行。

判断每一个字串是否是回文串，找出最长。

事件复杂度：O(n³)   (1000的输入， n³复杂度， 1000³=10⁹，会超1000ms）

 class Solution {
 public:
     string longestPalindrome(string s) {
         string res = "";
         int len = s.length();
         if(len ==  || len == )
             return s;

         for(int i=; i<len; i++)
         {
             string tmpStr = "";
             tmpStr += s[i];
             int j = i+;
             for(; j<len; j++)
             {
                 if(s[j] != s[i])
                     tmpStr += s[j];
                 else
                 {
                     tmpStr += s[j];
                     if(judge(tmpStr) && tmpStr.length() > res.length())
                         res = tmpStr;
                 }
             }
         }
         if(res.length() == )
             res += s[];
         return res;
     }

     bool judge(string s)
     {
         int len = s.length();
         for(int k=; k<=len/; k++)
             if(s[k] != s[len--k])
                 return false;
         return true;
     }
 };

思路2：DP

 class Solution {
 public:
     string longestPalindrome(string s) {
         int len = s.length();
         if(len < )
             return s;
         bool dp[len][len] = {};
         int start = , span = ;

         for(int i=; i<len; i++)
         {
             dp[i][i] = ;
             for(int j=; j<i; j++)
             {
                 dp[j][i] = (s[i]==s[j] && (i-j <  || dp[j+][i-]));
                 if(dp[j][i] && (i-j+ > span))
                 {
                     span = i-j+;
                     start = j;
                 }
             }
         }
         return s.substr(start, span);
     }
 };