leetcode个人题解——#5 Container with most water

class Solution {
public:
    string longestPalindrome(string s) {
        int length = s.length();
        if (length == ) return s;
        int len = ;
        int begin = ;
        int **pS = new int *[length];//palindromic state
        for (int i = ;i < length; i++)
             pS[i] = new int[length+];
        for( int i = ; i < length;i++)
        {
            pS[i][] = ;
            pS[i][] = ;
        }
        for(int i = ; i <= length; i++)
        {
            for (int j = ;j < length-i+; j++)
            {
                if (s[j] == s[j+i-] && pS[j+][i-] == ) {
                    len = i;
                    pS[j][i] = ;
                    begin = j;
                }
            }
        }
        string ans = "";
        for (int i = begin; i < begin + len; i++)
            ans += s[i];
        return ans;
        delete [] pS;
    }
};

对暴力解法做了一点优化，中间每一步结果存了下来，存放在pS[i][[j]中，这个数组意义是从第i个字符开始后长度为j的字符串是否为回文。

附上大佬的解法：

string longestPalindrome(string s) {
    if (s.empty()) return "";
    if (s.size() == ) return s;
    int min_start = , max_len = ;
    for (int i = ; i < s.size();) {
      if (s.size() - i <= max_len / ) break;
      int j = i, k = i;
      while (k < s.size()- && s[k+] == s[k]) ++k; // Skip duplicate characters.
      i = k+;
      while (k < s.size()- && j >  && s[k + ] == s[j - ]) { ++k; --j; } // Expand.
      int new_len = k - j + ;
      if (new_len > max_len) { min_start = j; max_len = new_len; }
    }
    return s.substr(min_start, max_len);
}