[LeetCode]5. Longest Palindromic Substring

Description

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.

Input: “cbbd”
Output: “bb”

Discussion

这道题可以采用动态规划的算法。我们维护一个二维数组dp。dp[i][j]表示从i到j是否是一个回文串。

递推式关系为：

如果dp[i+1][j-1] == 1 && s[i] == s[j]，则dp[i][j] = 1

算法的时间复杂度为O(n^2)。

C++ Code

class Solution {public:    string longestPalindrome(string s) {        int ** dp;        dp = new int * [s.length()];        //保存子串i～j是否是回文串        for(int i = 0; i < s.length(); i++)        {            dp[i] = new int[s.length()];        }        for(int i = 0; i < s.length(); i++)        {            for(int j = 0; j < s.length(); j++)            {                //把只有一个字符的串初始化为1，其他为0                if(i < j)                {                    dp[i][j] = 0;                }                else                {                    dp[i][j] = 1;                }            }        }        int maxLen = 0;        int answer_startLoc = 0;        int answer_endLoc = 0;        for(int len = 1; len < s.length(); len++)        {            for(int startLoc = 0; startLoc < s.length() - len; startLoc++)            {                //如果原子串是一个回文串，则在该子串前后添加两个相同的字符得到的新串也是回文串                if(s[startLoc] == s[startLoc + len] && dp[startLoc + 1][startLoc + len - 1] == 1)                {                    dp[startLoc][startLoc + len] = 1;                    if(len > maxLen)                    {                        maxLen = len;                        answer_startLoc = startLoc;                        answer_endLoc = startLoc + len;                    }                }            }        }        return s.substr(answer_startLoc, answer_endLoc - answer_startLoc + 1);    }};