Leetcode Palindrome Partitioning II

題目：

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.

解題：

思路類似Palindrome Partitioning，首先構造DP數組，然後再次dp，狀態轉移方程：

f[j] = min(f[j], f[i - 1] + 1);

f[i + 1] = min(f[i + 1], f[i] + 1);

代碼：

class Solution {
public:
    int minCut(string s) {
        makeDP(s);
        int f[SIZE];
        if(s == "") return 0;
        for(int i = 0; i <= s.size(); i ++)
            f[i] = INF;
        f[0] = 1;

        for(int i = 0; i < s.size(); i ++) {
            for(int j = i + 1; j < s.size(); j ++)
                if(dp[i][j]) {
                    if(i)
                        f[j] = min(f[j], f[i - 1] + 1);
                    else
                        f[j] = 1;
                }
            f[i + 1] = min(f[i + 1], f[i] + 1);
        }
        return f[s.size() - 1] - 1;
    }
private:
    static const int SIZE = 2000;
    static const int INF = 1 << 30;
    bool dp[SIZE][SIZE];

    void makeDP(string s) {
        memset(dp, 0, sizeof(dp));
        for(int i = s.size() - 1; i >= 0; i --)
            for(int j = i; j < s.size(); j ++) 
                if(s[i] == s[j] && (i == j || j == i + 1 || dp[i + 1][j - 1]))
                    dp[i][j] = 1;
    }
};