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-];

    }

};

巴特西

132. Palindrome Partitioning II (String; DP)

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