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.size();
        vector<int> cut(n+1, 0);  // number of cuts for the first k characters
        for (int i = 0; i <= n; i++) cut[i] = i-1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; i-j >= 0 && i+j < n && s[i-j]==s[i+j] ; j++) // odd length palindrome
                cut[i+j+1] = min(cut[i+j+1],1+cut[i-j]);

            for (int j = 1; i-j+1 >= 0 && i+j < n && s[i-j+1] == s[i+j]; j++) // even length palindrome
                cut[i+j+1] = min(cut[i+j+1],1+cut[i-j+1]);
        }
        return cut[n];
    }
};