【leetcode】948. Bag of Tokens

题目如下：

You have an initial power P, an initial score of 0 points, and a bag of tokens.

Each token can be used at most once, has a value token[i], and has potentially two ways to use it.

• If we have at least token[i] power, we may play the token face up, losing token[i] power, and gaining 1 point.
• If we have at least 1 point, we may play the token face down, gaining token[i] power, and losing 1 point.

Return the largest number of points we can have after playing any number of tokens.

Example 1:

Input: tokens = [100], P = 50
Output: 0

Example 2:

Input: tokens = [100,200], P = 150
Output: 1

Example 3:

Input: tokens = [100,200,300,400], P = 200
Output: 2

Note:

1. tokens.length <= 1000
2. 0 <= tokens[i] < 10000
3. 0 <= P < 10000

解题思路：本题可以用贪心算法。即可以得分的时候优先选择当前最小的token[i]得分，不能得分的时候选择当前可选的最大的token[i]换取power。所以只需要将tokens排序，分别从两端选择当前最小/最大的元素即可。

代码如下：

class Solution(object):
    def bagOfTokensScore(self, tokens, P):
        """
        :type tokens: List[int]
        :type P: int
        :rtype: int
        """
        tokens.sort()
        low = 0
        high = len(tokens) - 1
        res = 0
        point = 0
        while low <= high:
            if P >= tokens[low]:
                point += 1
                P -= tokens[low]
                low += 1
                res = max(res,point)
            elif point > 0:
                point -= 1
                P += tokens[high]
                high -= 1
            else:
                low += 1
        return res