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.

這道題是讓找到把原字串拆分成迴文串的最小切割數，需要用動態規劃Dynamic Programming來做，使用DP的核心是在於找出遞推公式，之前有道

class Solution {
public
 
:
    int minCut(string s) {
        int len = s.size();
        bool P[len][len];
        int dp[len + 1];
        for (int i = 0; i <= len; ++i) {
            dp[i] = len - i - 1;
        }
        for (int i = 0; i < len; ++i) {
            for (int j = 0; j < len; ++j) {
                P[i][j] 
 
= false;
            }
        }
        for (int i = len - 1; i >= 0; --i) {
            for (int j = i; j < len; ++j) {
                if (s[i] == s[j] && (j - i <= 1 || P[i + 1][j - 1])) {
                    P[i][j] = true;
                    dp[i] = min(dp[i], dp[j + 1] + 1);
                }
            }
        }
        return dp[0];
    }
};