动态编程?给定一个数组 A[0..n]，让 M(i) 成为使用索引为 0..i 的元素的最优解.然后M(-1) = 0(在循环中使用)，M(0) = A[0]，M(i) = max(M(i - 1), M(i - 2) + A[i]) 对于 i = 1, ..., n.M(n) 就是我们想要的解决方案.这是 O(n).您可以使用另一个数组来存储为每个子问题做出的选择，从而恢复实际选择的元素.

Dynamic programming? Given an array A[0..n], let M(i) be the optimal solution using the elements with indices 0..i. Then M(-1) = 0 (used in the recurrence), M(0) = A[0], and M(i) = max(M(i - 1), M(i - 2) + A[i]) for i = 1, ..., n. M(n) is the solution we want. This is O(n). You can use another array to store which choice is made for each subproblem, and so recover the actual elements chosen.