Leetcode 877 Stone Game

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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.

和leetcode 486题一样，只是更简单了
分析：

  1. 不妨认为Alex第一次拿的piles[0]，那么剩下的是piles[1:]
  2. Lee拿piles[1]还是piles[-1]不取决于这两个数哪个大，而是拿了某个数之后使得剩下的数中Alex能获得的总数小
  3. 若定义firstscore(piles)表示在当前piles数组中，先拿的人最后能获得的总分数，则alex的分数应该是max(piles[0]+min(firstscore(piles[1:-1]),firstscore(piles[2:])), piles[-1] + min(firstscore(piles[1:-1]), firstscore(piles[:-2])))

思路：

  1. 因为涉及到大量重复计算，所以需要采用记忆化的思想，用一个cache存已经计算过的值
  2. 但如果用piles作为key值，所花的时间和空间都太大了，所以改成用下标替代的方法
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def stoneGame(piles):
	cache = {}
	def firstscore(i,j):
		if i>=j: return 0
		if j==i+1 and j < len(piles): return piles[i]
		if (i,j) in cache: return cache[i,j]
		res = max(piles[i]+min(firstscore(i+2,j), firstscore(i+1,j-1)) , piles[j-1] + min(firstscore(i+1,j-1), firstscore(i,j-2)))
		cache[i,j] = res
		return res
	
	Alex = firstscore(0,len(piles))
	Lee = sum(piles) - Alex
	return Alex > Lee