877.Stone Game C++题解

题目

Alex and Lee play a game with piles of stones. There are an even 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. The total number of stones is odd, so there are no ties.

Alex and Lee take turns, with Alex starting first. Each turn, a player takes the entire pile of stones from either the beginning or the end of the row. This continues until there are no more piles left, at which point the person with the most stones wins.

Assuming Alex and Lee play optimally, return True if and only if Alex wins the game.

Example 1:

Input: [5,3,4,5]
Output: true
Explanation: 
Alex starts first, and can only take the first 5 or the last 5.
Say he takes the first 5, so that the row becomes [3, 4, 5].
If Lee takes 3, then the board is [4, 5], and Alex takes 5 to win with 10 points.
If Lee takes the last 5, then the board is [3, 4], and Alex takes 4 to win with 9 points.
This demonstrated that taking the first 5 was a winning move for Alex, so we return true.

Note:

1. 2 <= piles.length <= 500
2. piles.length is even.
3. 1 <= piles[i] <= 500
4. sum(piles) is odd.

解答一

考虑只有两堆石头的情况，显然存在着必胜策略。

再考虑有四堆时候的情况，不妨设为$abcd$；则Alex一定能够取得$bd$ 或者 $ac$，所以只要选择这两者中大的情况就也存在必胜策略。

再考虑当 N > 4 的情况，Alex总是可以获得第奇数堆（第一次选择第一堆，然后每次选择与对手的选择相邻的一堆），同理 也可以获得第偶数堆，所以和四堆的情况相似，只要选择两者中大的情况就存在必胜策略。

代码如下：

class Solution {
public:
    bool stoneGame(vector<int>& piles) {
        return true;
    }
};