5. Longest Palindromic Substring

背景

“回文”是指正读反读都能读通的句子，它是古今中外都有的一种修辞方式和文字游戏，如“我为人人，人人为我”等。在数学中也有这样一类数字有这样的特征，成为回文数（palindrome number）。

设n是一任意自然数。若将n的各位数字反向排列所得自然数n1与n相等，则称n为一回文数。例如，若n=1234321，则称n为一回文数；但若n=1234567，则n不是回文数。
注意：
1. 偶数个的数字也有回文数124421
2. 小数没有回文数

题目

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.

Example:

Input: “cbbd”

Output: “bb”

代码实现

public string LongestPalindrome(string s)        {            //considering: adccbaabc. we get the palindrome "cc" and "cbaabc"            //How do we get the "cbaabc"?             //first k points to second 'a' and j points to first 'b'  by such logics:             //while (k < s.Length - 1 && s[k + 1] == s[k])            //    ++k; // moves when happing duplicate characters.            //then we compare the k+1 index to j-1 index, such s[k+1]== s[j-1] is ok. It's finished by expand logics:            //while (k < s.Length - 1 && j >= 1 && s[k + 1] == s[j - 1])            //{            //    ++k;            //    --j;            //}            if (s.Length== 0 || s.Length==1) return s;            int i=0, minstart = 0, maxlen = 1;            while(i<s.Length){                if (s.Length - i <= maxlen / 2) break;                int j = i, k = i;                while (k < s.Length - 1 && s[k + 1] == s[k])                     ++k;                 i = k + 1;                while (k < s.Length - 1 && j >= 1 && s[k + 1] == s[j - 1]) {                    ++k;                    --j;                }                int newlen = k - j + 1;                if (newlen > maxlen){                    minstart = j;                    maxlen = newlen;                }            }            return s.Substring(minstart, maxlen);        }

题库

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完成题目索引

http://blog.csdn.net/daigualu/article/details/73008186