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.

class Solution {public:    int minCut(string s) {        int n = s.length();        if (n < 1)        {            return 0;        }        bool isPal[n][n];//记录从i到j之间的字符串是否为回文        memset(isPal, false, n*n*sizeof(bool));        int cut[n];//记录从0到i的字符串的最小切割数量        cut[0] = 0;        for (int i = 1; i < n; i++)        {            cut[i] = cut[i-1] + 1;            for (int j = 0; j <= i; j++)            {                if (s[i] == s[j] && (i-j < 2 || isPal[j+1][i-1]))                {                    isPal[j][i] = true;                    if (j == 0)                    {                        cut[i] = 0;                    }                    else if (cut[j-1]+1 < cut[i])                    {                        cut[i] = cut[j-1] + 1;                    }                }            }        }        return cut[n-1];    }};