LeetCode #132 - Palindrome Partitioning II

题目描述：

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.

和Palindrome Partitioning类似，但是要求最小拆分次数。动态规划的思路也是一样的，但是不需要求拆分的子字符串集合，而是保留最小的拆分次数。

class Solution {
public:
    int minCut(string s) {
        int n=s.size();
        vector<bool> row(n,false);
        vector<vector<bool>> isPalin(n,row);
        vector<int> min_cut(n,INT_MAX);
        
        if(n==0) return 0;
        else if(n>=1) min_cut[0]=0;
        for(int j=1;j<n;j++)
        {
            for(int i=0;i<=j;i++)
            {
                if(i==j) isPalin[i][j]=true;
                else if(j==i+1&&s[i]==s[j]) isPalin[i][j]=true;
                else if(isPalin[i+1][j-1]&&s[i]==s[j]) isPalin[i][j]=true;
                if(isPalin[i][j]==true)
                {
                    if(i==0) min_cut[j]=0;
                    else min_cut[j]=min(min_cut[j],min_cut[i-1]+1);
                }
            }
        }
        return min_cut[n-1];
    }
};