2 Player and N Coin

class Solution {

    public void printChoices(int[] A, int[][] C) {
        System.out.println("硬币列表如下:");
        for(int i=0,l=A.length; i<l; i++) {
            System.out.print(A[i] + "\t");
        }
        System.out.println("");
        final int N = C.length;
        String[] names = {"本人", "对手"};
        int L = 0, R = N-1, WHO = 0;
        while(L<=R) {
            int c = C[L][R];
            System.out.println(names[WHO] + "选择第 " + c + " 个元素" + A[c]);
            WHO = (WHO+1)%2;
            if(L==c) {
                L = L+1;
            }
            if(R==c) {
                R = R-1;
            }
            if(L == R) {
                break;
            }
        }
    }

    public int maxMoney(int[] A) {
        final int N = A.length;
        int[][] C = new int[N][N];
        int[][] M = new int[N][N]; // 每个子问题最优解（赚的最多钱）

        // 设M(i,j) 为从第i到第j个硬币中能获得的最大收入
        // j-i>=2 时
        // M(i,j) = max(
        //      A[i] + min(M(i+2,j), M(i+1,j-1)),
        //      A[j] + min(M(i+1,j-1), M(i,j-2)),
        // );

        // 初始化坐标 [i,i] 和 [i,i+1] 处的决策和最大收入
        for(int i=0; i<N; i++) {
            M[i][i] = A[i];
            C[i][i] = i;
            if(i<N - 1) {
                M[i][i+1] = Integer.max(A[i], A[i+1]);
                C[i][i+1] = (A[i] < A[i+1] ? i+1 : i);
            }
        }

        // 计算所有 [i,i+2] [i,i+3]... 处的决策和最大收入
        for(int i=N-2; i>=0; i--) {
            for(int j=i+2; j<N; j++) {
                int I = A[i] + Integer.min(M[i+2][j], M[i+1][j-1]);
                int J = A[j] + Integer.min(M[i+1][j-1], M[i][j-2]);
                M[i][j] = Integer.max(I, J);
                C[i][j] = I>J ? i : j;
            }
        }

        printChoices(A, C);
        return M[0][N-1];
    }

    public static void main(String[] args) {
        Solution s = new Solution();
        int[] a = {1,20,10,2};
        int r = s.maxMoney(a);
        System.out.println(r);
    }
}