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.

这道题很惭愧，想直接在上一道题的代码上改，结果还是重写了。

之前为了方便算回文，插入“#”。但是程序会报内存不足，所以不能用这种扩展方式，只能拿string.size来建dp数组

后面根据数组，用dp继续做cut。cut的时候 cut j = min（cut j ， cut i +1  （if dp[i+1][j] == 1））.要注意该推导式不需要dp[0][i] == 1

class Solution {
public:
    int minCut(string s) {
       vector<vector <int> >dp;
        int len = s.size();
        if(len == 1)return 0;

        for(int i = 0 ; i < len ;i++)
        {
            vector<int> tp(len,0);
            dp.push_back(tp);
        }
        for(int i  = 0  ; i < len ; i++)
        dp[i][i] = 1;
        
        for(int i = 0  ; i < len ;i++)
        {
            int j = 1;
            while(i-j>=0 && i+j<len && s[i-j] == s[i+j])
            {
                dp[i-j][i+j] = 1;
                j++;
            }
            int x = i , y =i+1;
            while(x>=0 && y<len)
            {
                if(s[x]==s[y])
                {
                    dp[x][y]=1;
                    x--;
                    y++;
                }
                else
                break;
            }
        }
        int num = 0;
        if(dp[0][len-1]==1)return 0;
        vector<int> cut(len ,1);
        for(int i = 0 ; i < len ;i++)
        if(dp[0][i]==1)cut[i] = 1;
        else
        cut[i] = i+1;
        
        for(int i = 1;i<len;i++)
        {
            for(int j = 0 ;j <i ;j++)
            {
                
                if(j+1<len && dp[j+1][i]==1)
                {
                    cut[i] = min(cut[j]+1,cut[i]);
                }
            }
        }
       return cut[len-1]-1;
    }
};