【leetcode】877. Stone Game

题目如下：

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.

解题思路：这类博弈问题是我的弱项，本题我参考了很多高手的答案才得到动态规划的状态转移方程。记dp[i][j]为piles[i][j]区间内先手可以赢后手的点数，假设当前dp[i][j]是Alex先手，所有Alex可以选择的石头是piles[i]或者piles[j]，如果Alex选择是piles[i]，那么区间piles[i+1][j]就对应Lee的先手，dp[i+1][j] 对应着Lee赢Alex的点数；当然如果Alex选择的是piles[j]，其实也是一样的，只不过下一手变成piles[i][j+1]。综合这两种情况，可以得出 dp[i][j] = max(piles[i] - dp[i+1][j] , piles[j] - dp[i][j-1]) 。

代码如下：

class Solution(object):
    def stoneGame(self, piles):
        """
        :type piles: List[int]
        :rtype: bool
        """
        dp = []
        for i in range(len(piles)):
            dp.append([0] * len(piles))
            dp[i][i] = piles[i]

        # 这里的计算逻辑是j为inx，i为每一段石头的个数
        for i in range(1,len(dp)):
            for j in range(len(dp) - i):
                dp[j][j+i] = max(piles[j] - dp[j+1][j+i], piles[j+i] - dp[j][j+i-1])
        return dp[0][-1] > 0