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Leetcode-Palindrome Partitioning II

时间:2014-12-12 06:41:14      阅读:267      评论:0      收藏:0      [点我收藏+]

Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

For example, given s = "aab",
Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.

Analysis:

DP

 

Solution:

 1 public class Solution {
 2     public int minCut(String s) {
 3         if (s.length()==0) return 0;
 4 
 5         int[] minCut = new int[s.length()+1];
 6         minCut[0] = -1;
 7         boolean[][] valid = new boolean[s.length()][s.length()];        
 8         for (int i=0;i<s.length();i++){
 9             valid[i][i]=true;
10         }        
11 
12         for (int i=1;i<=s.length();i++){
13             minCut[i] = minCut[i-1]+1;
14             for (int j=i-2;j>=0;j--)
15                 if (s.charAt(j)==s.charAt(i-1) && (j==i-2 || valid[j+1][i-2])){
16                     valid[j][i-1]=true;
17                     minCut[i] = Math.min(minCut[i],minCut[j]+1);
18                 }
19                 
20         }
21 
22         return minCut[s.length()];
23     }
24 
25 }

 

Leetcode-Palindrome Partitioning II

原文:http://www.cnblogs.com/lishiblog/p/4158926.html

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