[LeetCode]5 Longest Palindromic Substring(C++，Python实现)，

[LeetCode]5 Longest Palindromic Substring(C++，Python实现)，

LeetCode OJ的第五题，如果有问题或者给我指点欢迎来信讨论ms08.shiroh@gmail.com

题目描述

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.
寻找一个字符串的最大回文子串

思路

最简单直观的想法就是一个一个子串试过去是否为回文。但是这样的时间复杂度很高。而且其实有很多不必判断的过程，如i到j之间，如果i-1与j-1已经不相同了，那么判断i,j毫无意义。所以很简单的思路是判断每个点扩展的最大串。遇到相同的字母相连的情况可以合在一起判断。这种方法在最坏的情况下复杂度会达到O(n^2)。 有大神给出了O(n)时间复杂度的算法，具体可以看http://leetcode.com/2011/11/longest-palindromic-substring-part-ii.html

代码

Python O(n^2)

class Solution:
    # @return a string
    def longestPalindrome(self, s):
        if not s or len(s) == 1:
            return s
        # record the result value
        max_length = 1
        start_index = 0
        end_index = 0
        for i in range(0, len(s)-1):
            count = 1
            # aba
            if s[i] != s[i+1]:
                while i-count >= 0 and i + count < len(s) and s[i-count] == s[i+count]:
                    count += 1
                if (count-1) * 2 + 1 > max_length:
                    max_length = (count-1) * 2 + 1
                    start_index = i - count + 1
                    end_index = i + count - 1
            # xaaaaax
            else:
                count_repeat = 1
                count = 0
                while i+1 < len(s) and s[i] == s[i+1]:
                    i += 1
                    count_repeat += 1
                while i-count_repeat+1-count >= 0 and i + count < len(s) and s[i-count_repeat+1-count] == s[i+count]:
                    count += 1
                if (count-1) * 2 + count_repeat > max_length:
                    max_length = (count-1) * 2 + count_repeat
                    start_index = i - count -count_repeat + 2
                    end_index = i + count - 1
        return s[start_index:end_index+1]

C++ O(n^2)

class Solution {
public:
    string longestPalindrome(string s) {
        if (s == "" || s.size() == 1)
		return s;
	// 记录结果
	int max_length = 0, start_index = 0;
	for (int i = 0; i < s.size()-1; ++i){
	int count = 1;
	// 形如aba的回文
	if (s[i] != s[i+1]){
	    while (i-count >= 0 && i+count < s.size() && s[i-count] ==s[i+count])
		++count;
	    if ((count-1) * 2 + 1 > max_length){
		max_length = (count-1) * 2 + 1;
		start_index = i - count + 1;
	    }
	} else { // 形如xaaaaax这样的回文
	    int count_repeat = 0;
	    count = 0;
	    while (i+1 < s.size() && s[i] == s[i+1]){
		++i;
		++count_repeat;
	    }
	    while (i-count_repeat-count >= 0 && i+count < s.size() && s[i-count_repeat-count] ==s[i+count])
		++count;
	        if ((count-1) * 2 + 1 + count_repeat > max_length){
		    max_length = (count-1) * 2 + 1 + count_repeat;
		    start_index = i - count + 1 - count_repeat;
		}
	    }
	}
	return s.substr(start_index, max_length);
    }
};