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.

1:注意特殊情况；2:从前到后遍历字符串，在遍历的过程中检测从i到j之间的字符子串是不是回文，如果是回文，设置回文标识，并计算相应的最小分割。

        int minCut(string s)
    {
        if(s.length() <= 1)
        {
            return 0;
        }
        
        int size = (int)s.length();
        vector<int> minResult(size + 1, 0);
        for(int i = 0; i <= size; i++)
        {
            minResult[i] = i  - 1;
        }
        
        vector<vector<bool> > flagResult(size, vector<bool>(size, false));
        
        for(int i = 0; i < size; i++)
        {
            for(int j = 0; j <= i; j++)
            {
                if(s[i] == s[j] && (i - j <= 2 || flagResult[j+1][i-1] == true))
                {
                    flagResult[j][i] = true;
                    minResult[i+1] = (minResult[i+1] > minResult[j] + 1 ? minResult[j] + 1 : minResult[i+1]);
                }
            }
        }
        return minResult[size];
    }