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.

一刷
题解：
初始定义一个一维数组minCuts，用来表示s.substring(0, i + 1)时的最小cut数。

再定义一个二维布尔数组isPalindrome，用来表示 s.substring(j, i)是否是palindrome。

核心算法是，假如isPalindrome[j][i]是palindrome，说明 j - 1至i + 1只需要1个cut， 因此对每一个i， minCuts[i]可以进行更新 - 比较现有 minCuts[i] 与 minCuts[j - 1] + 1。 这里也是一个一维的dp。

public class Solution {
    public int minCut(String s) {
        int len = s.length();
        int[] minCuts = new int[len]; //minCuts[i] is min cut for s.substring(0, i+1)
        boolean[][] isPalindrome = new boolean[len][len];
        
        for(int i=0; i<len; i++){
            minCuts[i] = Integer.MAX_VALUE;
            
            for(int j=0; j<=i; j++){
                if(s.charAt(j) == s.charAt(i)){//s.substring(j, i+1) is palindrome
                    if(i-j <=1 || isPalindrome[j+1][i-1]){
                        isPalindrome[j][i] = true;
                        if(j==0) minCuts[i] = 0;
                        else minCuts[i] = Math.min(minCuts[i], minCuts[j-1]+1);
                    }
                }
            }
        }
        
        return minCuts[len-1];
    }
}

二刷
思路同上

public class Solution {
    public int minCut(String s) {
        int len = s.length();
        int[] minCuts = new int[len];
        Arrays.fill(minCuts, Integer.MAX_VALUE);
        boolean[][] isPalin = new boolean[len][len];
        
        for(int i=0; i<len; i++){
            for(int j=0; j<=i; j++){
                if(s.charAt(i) == s.charAt(j)){
                    if(i-j<=1 || isPalin[j+1][i-1]){
                        isPalin[j][i] = true;
                        if(j == 0) minCuts[i] = 0;
                        else minCuts[i] = Math.min(minCuts[i], minCuts[j-1]+1);
                    }
                }
            }
        }
        return minCuts[len-1];
    }
}