[Dynamic Programming leetcode 132 Palindrome Partitioning II

problem:https://leetcode.com/problems/palindrome-partitioning-ii/

        Climb stairs type of problem. First calculate all possible palindromic sequence, DP [i] i represents the minimum divided front characters, strings found in all i palindromic ends, whichever is the smallest division that as a result.

class Solution {
public:
    int minCut(string s) {
        int n = s.size();
        vector<vector<bool>> palindrome(n, vector<bool>(n, false));
        for(int i = 0;i < n;i++)
        {
            palindrome[i][i] = true;
            for(int times = 0; times < 2; times++)
            {
                int j;
                if(times == 0) j = i - 1;
                else j = i;
                int k = i + 1;
                while(j >= 0 && k < n)
                {
                    if(s[j] == s[k])
                    {
                        palindrome[j][k] = true;
                        j--;
                        k++;
                    }
                    else break;
                }
            }
        }
        
        vector<int> dp(n + 1);
        dp[0] = -1;
        for(int i = 0;i < n; i++)
        {
            dp[i + 1] = i;
            for(int j = 0;j <= i; j++)
            {
                if(palindrome[j][i])
                {
                    dp[i + 1] = min(dp[i + 1], dp[j] + 1);
                }
            }
        }
        return dp[n];
    }
};