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Leetcode Palindrome Partitioning II

2015-04-21 00:21 281 查看
题目地址:https://leetcode.com/problems/palindrome-partitioning-ii/

题目解析:此问题可以使用动态规划,用一个数组保存前i个字符需要的最少cut数,前i+1个字符串的最小cut数为前j个字符所需的cut数(j+1到i个字符为回文)+1;

题目解答:

public class Solution {
public int minCut(String s) {
if(s == null || s.length() == 0){
return 0;
}

int[] cutNum = new int[s.length()+1];
boolean[][] palindromeMap = new boolean[s.length()][s.length()];
cutNum[0] = -1;
for(int i=1;i<=s.length();i++){
cutNum[i] = i-1;
for(int j=0;j<=i-1;j++){
palindromeMap[j][i-1] = false;
if(s.charAt(j) == s.charAt(i-1) && (i-1-j<=2 || palindromeMap[j+1][i-2])){
palindromeMap[j][i-1] = true;
cutNum[i] = Math.min(cutNum[i], cutNum[j]+1);
}
}
}
return cutNum[s.length()];
}
}
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