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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.
很经典的DP问题：
算法分析：
思路1：最土的算法，将所有的子串求出来，再选出最长的回文子串。不用试，肯定超时。
思路2：挑出最短的回文串（单个字符，或者两个相邻的相同字符），然后向两端辐射，时间复杂度O(n^2)，空间复杂度可以优化到O(1)【我木有优化= =】
代码如下：
1 public String longestPalindrome(String s) { 2 if(s == null || s.length() == 0) return ""; 3 String maxsub = ""; 4 for(int i = 0; i < s.length(); i++){ 5 String str = getPalindrome(s, i, i,s.length()); 6 if(str.length() > maxsub.length()) 7 maxsub = str; 8 if(i < s.length() - 1 && s.charAt(i) == s.charAt(i + 1)){ 9 str = getPalindrome(s, i, i + 1,s.length()); 10 if(str.length() > maxsub.length()) 11 maxsub = str; 12 } 13 } 14 return maxsub; 15 } 16 private String getPalindrome(String s,int begin,int end,int prime){ 17 while(begin >= 0 && end < s.length()){ 18 if(s.charAt(begin) == s.charAt(end)){ 19 begin--; 20 end++; 21 }else{ 22 break; 23 } 24 } 25 return s.substring(begin+1, end); 26 }
思路3：DP
参考同学的做法：
dp[i][j] 代表从i到j的子串是否是palindrome。自下而上自左而右计算dp数组。时空复杂度都是 O(n^2)。
dp[i][j]=1 if:
- i=j；
- s.charAt(i)==s.charAt(j) && j-i<2
- s.charAt(i)==s.charAt(j) && dp[i+1][j-1]==1
1 public class Solution { 2 public String longestPalindrome(String s) { 3 if(s == null || s.length() == 0) return ""; 4 boolean[][] dp = new boolean[s.length()][s.length()]; 5 int start = 0, end = 0,len = 0; 6 for(int i = s.length() - 1; i >= 0; i--){ 7 for(int j = i; j < s.length(); j++){ 8 if(i == j || (s.charAt(i) == s.charAt(j) && j - i == 1) || (s.charAt(i) == s.charAt(j) && dp[i + 1][j - 1])){ 9 dp[i][j] = true; 10 if(j - i + 1 > len){ 11 len = j - i + 1; 12 start = i; 13 end = j; 14 } 15 } 16 } 17 } 18 return s.substring(start, end + 1); 19 } 20 }
[leetcode]Longest Palindromic Substring,布布扣,bubuko.com
[leetcode]Longest Palindromic Substring
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原文地址：http://www.cnblogs.com/huntfor/p/3883871.html
