leetcode -- 877. Stone Game

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:

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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/stone-game
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

 

 

 

coder_hezi思路：

1. dp[i][j]表示 从i堆石子到j堆石子，Alex能赢li多少分，如果到最后，这个值是正值则说明Alex会赢

2. dp[i][j]取决于：

如果Alex选择了第i堆，则她领先的分数是 第i堆- 剩下i+1堆到i+j堆她能赢的分数(对手会选的）

如果Alex选择了最后一堆j，则就是最后一堆(i+j) - 第i堆到（i+j-1）堆能赢的(对手会选的）

3. 边界：dp[i][i] = piles[i]，只有一堆就是ALex赢

4. 最后返回第0堆到最后一堆，Alex的分数是否是正值

class Solution {
public:
    bool stoneGame(vector& piles) {
        int n = piles.size();
        int dp[n][n];
        for(int i = 0; i < n; i++){
            dp[i][i] = piles[i];
        }
        for(int j = 1; j < n; j++){
            for(int i = 0; i < n-j; i++){
                dp[i][i+j] = max(piles[i]-dp[i+1][i+j], piles[i+j]-dp[i][j+i-1]);
            }
        }
        return dp[0][n-1] > 0;
    }
};