【LeetCode】 Longest Palindrome Substring

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题目链接：http://oj.leetcode.com/problems/longest-palindromic-substring/

题目描述：

Longest Palindromic Substring

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.

题目意思：找出给定字符串中最常的回文字串，输入保证有且仅有一个正确结果。

解法一：

暴力求解，时间复杂度O(n^2).提交会超时。

代码如下：

string longestPalindrome(string s) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        int result = 0;    int start = -1;    int len = s.length();    if(len <= 0) return NULL;    if(len == 1) return s;    for(int i=0;i < len; ++i){    int j = i+1;    while(j < len){    if( isPalindrome(s, i, j)){    int temp = j - i + 1;    if(temp > result){    result = temp;    start = i;    }    }    j++;    }    }    return s.substr(start, result);    }        bool isPalindrome(string s, int start, int end){        while(start < end){            if(s[start++] != s[end--])                return false;        }        return true;    }

解法二：

暴力求解，时间复杂度O(n^2)。能通过，rp好吗!

思路：分别以每个位置为中心向两边扩展。

代码如下:

string longestPalindrome(string s) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.    int result = 0;    int start = 0;    int len = s.length();    if(len <= 0) return NULL;    if(len == 1) return s;    for(int i=0;i < len; ++i){    int f = i+1,b = i;    int temp;    while(b >=0 && f < len && s[b] == s[f])     b--, f++;    temp = f - b - 1;    if(temp > result){    result = temp;    start =  b + 1;    }        f = i+1, b = i-1;    while(b >=0 && f < len && s[b] == s[f])     b--, f++;    temp = f - b - 1 ;    if(temp > result){    result = temp;    start =  b + 1;    }    }    return s.substr(start, result);    }

未完待续…………DP和O(n)

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