[LeetCode]5. Longest Palindromic Substring

  "求一个字符串的最长回文子串"

Posted by Stephen.Ri on October 1, 2017

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);
    }
};