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:
1 | Input: [5,3,4,5] |
Note:
2 <= piles.length <= 500
piles.length
is even.1 <= piles[i] <= 500
sum(piles)
is odd.
分析
$dp[i][j]$ 表示 $piles[i]~piles[j]$石头堆中,Alex 最多可以赢 Lee 的数量,则有 $dp[i][j] = max(piles[i] - dp[i+1][j], piles[j] - dp[i][j-1])$,其中如果 Alex 取第一块,$dp[i+1][j]$ 为 Lee 赢 Alex 的数量,因此 $piles[i] - dp[i+1][j]$ 为 Alex 可以赢 Lee 的数量。
解决方案1(Java)
1 | class Solution { |
解决方案2(Java)
题目中其实有个隐含条件,石头的堆树为偶数,这确保了先取的 Alex 可以取1,3,5… 2n+1 或 2,4,6,2n,而总数量为奇数,这两个序列的和总有一个更大,所以先取的 Alex 一定可以赢。
1 | class Solution { |