LeetCode - 5 Longest Palindromic Substring 最长回文子字符串
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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.
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初始条件:
1 单个字符是回文
2 紧挨着的两个相同字符也是回文
3 如果两端字符相同,取决于内部子串情况
public class Solution { public string LongestPalindrome(string s) { if (s == string.Empty || s.Length == 1) return s; int n = s.Length; int longestBegin = 0; int maxLen = 1;//单个字符是回文 bool[,] dp = new bool[n, n]; char[] sArray = s.ToCharArray(); //单个字符是回文,正序循环 for (int i = 0; i < n - 1; i++) dp[i, i] = true; //紧挨着的两个相同字串也是回文,正序循环 for (int i = 0; i < n - 1; i++) { if (sArray[i] == sArray[i + 1]) { dp[i, i + 1] = true; longestBegin = i; maxLen = 2; } } //考虑到dp的情况,必须从小到大; 所以倒序循环; for (int i = n - 3; i >= 0; i--) for (int j = i + 2; j < n; j++) if(sArray[i]== sArray[j]&&dp[i+1,j-1])//如果两短字符串相同,则取决于内部子串情况 { dp[i, j] = true; if(j-i+1>maxLen)//如果大于最长的,再做修改 { longestBegin = i; maxLen = j - i + 1; } } //截取回文 return s.Substring(longestBegin, maxLen); }}
注意问题:
考虑到dp[]数组内的内容,最后一遍的循环,需要长度从小到大跑,不能从i,j跨度从大到小;否则就会报错;
原因:
if(sArray[i]== sArray[j]&&dp[i+1,j-1])此处无法进入后半部分的分支;
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