132. Palindrome Partitioning II（js）

132. Palindrome Partitioning II

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.

Example:

Input: "aab"
Output: 1
Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.
题意：给定一个字符串，问最少切割多少次使得每个子串都是回文
代码如下：

/**
 * @param {string} s
 * @return {number}
 */
var minCut = function(s) {
    let arr=s.split('');
    let n=s.length;
    let pal=[];
    let cut=[];
    for(let i=0;i<n;i++){
        cut[i]=0;
    }
    for(let i=0;i<n;i++){
        pal[i]=[]
        for(let j=0;j<n;j++){
            pal[i][j]=false;
        }
    }
    
    for(let i=0;i<n;i++){
        let min=i;
        for(let j=0;j<=i;j++){
            if((arr[i]===arr[j]) && (j+1>i-1 || pal[j+1][i-1])){
                pal[j][i]=true;
                min=j===0?0:Math.min(min,cut[j-1]+1);
            }
        }
        cut[i]=min;
    }
    return cut[n-1];
};