132. Palindrome Partitioning II (String; DP)

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.

思路: 除了用dp[i][j]记录从i到j是否是Palindrome，还要用cut[i]存储到i位置位置最少的cut数

class Solution {
public:
    int minCut(string s) {
        int len = s.length();
        vector<vector<bool>> dp(len, vector<bool>(len, false));
        vector<int> cut(len,);
        for(int i = ; i < len; i++) dp[i][i]=true;
        for(int i = ; i < len; i++){
            cut[i]=cut[i-]+;
            for(int j = ; j < i; j++){ //traverse the length
                if(s[i]==s[j]){
                    if(j==i-) dp[j][i] = true;
                    else dp[j][i]=dp[j+][i-];
                    if(dp[j][i]){
                        if(j==) cut[i]=;
                        else cut[i]=min(cut[j-]+, cut[i]);
                    }
                }
            }
        }
        return cut[len-];
    }
};