LeetCode_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.

For example, given s = "aab",
Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.

  分析: D[i] = 区间[i,n)之间最小的cut数,n为字符串长度, 则,D[i] = min(D[i],1+D[j+1]) i<=j <n

                P[i][j] = str[i] == str[j] && P[i+1][j-1];

class Solution {
public:
    int minCut(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int len = s.size();
        if(len < 2) return 0 ;
        vector<vector<bool>> flag(len, vector<bool>(len, false)) ;
        vector<int> dp(len+1);
        
        for(int i = 0; i<= len ; i++)
            dp[i] = len - i;
            
        for(int i = len -1 ; i >= 0 ; i--)
             for(int j = i; j < len ; j++)
                 if(s[i] == s[j] &&(j-i< 2 || flag[i+1][j-1] ) )
                 {
                        flag[i][j] = true;
                        dp[i] = dp[i] < dp[j+1] +1 ? dp[i]  : dp[j+1] +1 ;
                 }
        
            
        return dp[0]-1;            
    }
};
原文地址:https://www.cnblogs.com/graph/p/3223942.html