Leetcode Palindrome Partitioning II

 

题目地址：https://leetcode.com/problems/palindrome-partitioning-ii/

题目解析：此问题可使用动态规划，用一个数组保存前i个字符须要的最少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()];
    }
}