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);
     }
 }