Stone Game - 阿振的个人主页

Stone Game

题目描述

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.

算法分析

每次都是Alex先取，从前后两个中选取最大的，然后Lee再取，Lee再从前后两个中取最大的

这告诉我们什么，毫无悬念，Alex肯定会赢呀

C++实现

直接返回true就好了

bool stoneGame(vector<int>& piles) {
    return true;
}

模拟他们玩游戏的过程，使用两个指针分别代表指向最前面和最后面，依次移动指针

bool stoneGame(vector<int>& piles) {
    const size_t count = piles.size();
    vector<int> sum(count + 2, 0);
    for (size_t i = 0, j = count - 1, k = 2; i != j ; k++) {
        if (piles[i] > piles[j]) {
              sum[k] = piles[i++]  +  sum[k - 2];
        } else {
            sum[k] = piles[j--] + sum[k - 2];
        }
    }
    return sum[count] > sum[count + 1];
}