Stone Game II

Alex and Lee continue their games with piles of stones.  There are a number of piles arranged in a row, and each pile has a positive integer number of stones piles[i].  The objective of the game is to end with the most stones. 

Alex and Lee take turns, with Alex starting first.  Initially, M = 1.

On each player's turn, that player can take all the stones in the first X remaining piles, where 1 <= X <= 2M.  Then, we set M = max(M, X).

The game continues until all the stones have been taken.

Assuming Alex and Lee play optimally, return the maximum number of stones Alex can get.

Example 1:

Input: piles = [2,7,9,4,4]
Output: 10
Explanation:  If Alex takes one pile at the beginning, Lee takes two piles, then Alex takes 2 piles again. Alex can get 2 + 4 + 4 = 10 piles in total. If Alex takes two piles at the beginning, then Lee can take all three piles left. In this case, Alex get 2 + 7 = 9 piles in total. So we return 10 since it's larger. 

Constraints:

• 1 <= piles.length <= 100
• 1 <= piles[i] <= 10 ^ 4

思路：这题用记忆化递归做，比较好做；score代表的物理意义是以i为起点，我面对i ~  n - 1个石头的，我能够取到的最大值，也就是相对最大值，后面的sum 减去对手能够得到的最大值；这里有一维参数M，每次传下去的是Max(M, x) 因为下次我可以取[1,2*M]个数；

class Solution {
    public int stoneGameII(int[] A) {
        if(A == null || A.length == 0) {
            return 0;
        }
        int n = A.length;
        int[] sum = new int[n]; // sum 代表是：从后往前计算，i ~ n -1的和，因为我是求后面的和；
        sum[n - 1] = A[n - 1];
        for(int i = n - 2; i >= 0; i--) {
            sum[i] = sum[i + 1] + A[i];
        }
        
        int[][] cache = new int[n][n];
        return score(A, 0, 1, sum, cache);
    }
    
    private int score(int[] A, int i, int M, int[] sum, int[][] cache) {
        if(i >= A.length) {
            return 0;
        }
        if(i + 2*M >= A.length) {
            return sum[i];
        }
        if(cache[i][M] != 0) {
            return cache[i][M];
        }
        int min = Integer.MAX_VALUE; // 对手能够拿到的最小值；
        for(int x = 1; x <= 2*M; x++) {
            // 注意这里是i + x,因为i是以0开始的index，取x个，那么也包含i,所以第x + 1个，是对手取的，那么开始就是i + x;
            min = Math.min(min, score(A, i + x, Math.max(M, x), sum, cache));
        }
        cache[i][M] = sum[i] - min; // 我能拿到的最大值，就是减去对手所有可能性中最小的，那么就是我最大；
        return cache[i][M];
    }
}