[LeetCode] Longest Palindromic Substring

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本文介绍了两种高效算法来寻找给定字符串中的最长回文子串。第一种算法使用动态规划的方法，通过构建一个布尔型表格来记录字符串中各部分是否为回文。第二种算法采用中心扩展法，分别考虑奇数长度和偶数长度的回文情况。

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.

Solution 1:

class Solution {
public:
    string longestPalindrome(string s) {
        int n = s.length();
        int longestBegin = 0;
        int maxLen = 1;
        bool table[1000][1000] = {false};
        for(int i = 0; i < n; i++)
        {
            table[i][i] = true;
        }
        for(int i = 0; i < n-1; i++)
        {
            if(s[i] == s[i+1])
            {
                table[i][i+1] = true;
                longestBegin = i;
                maxLen = 2;
            }
        }
        for(int len = 3; len <= n; len++)
        {
            for(int i = 0; i < n-len+1; i++)
            {
                int j = i+len-1;
                if(s[i] == s[j] && table[i+1][j-1])
                {
                    table[i][j] = true;
                    longestBegin = i;
                    maxLen = len;
                }
            }
        }
        return s.substr(longestBegin, maxLen);
    }
};

Solution 2:

class Solution {
public:
    string expandAroundCenter(string s, int c1, int c2)
    {
        int l = c1, r = c2;
        int n = s.length();
        while(l >= 0 && r <= n-1 && s[l] == s[r])
        {
            l--;
            r++;
        }
        return s.substr(l+1, r-l-1);
    }
    string longestPalindrome(string s) {
        int n = s.length();
        if(n==0)    return "";
        string longest = s.substr(0,1);
        for(int i = 0; i < n-1; i++)
        {
            string p1 = expandAroundCenter(s,i,i);
            if(p1.length() > longest.length())
                longest = p1;
            string p2 = expandAroundCenter(s,i,i+1);
            if(p2.length() > longest.length())
                longest = p2;
        }
        return longest;
    }
};