LeetCode知识点总结 - 948

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于 2024-01-25 15:35:58 发布

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文章介绍了如何使用贪心算法解决一个关于BagofTokens的游戏问题，玩家通过合理选择面对上或面对下每个token来最大化得分，同时考虑当前的power和score。关键在于对tokens进行排序并维护两个指针，交替使用最小的token换取分数或用多余的分数换取最大power。

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948. Bag of Tokens

考点	难度
Greedy	Easy

题目

You have an initial power of power, an initial score of 0, and a bag of tokens where tokens[i] is the value of the ith token (0-indexed).

Your goal is to maximize your total score by potentially playing each token in one of two ways:

If your current power is at least tokens[i], you may play the ith token face up, losing tokens[i] power and gaining 1 score.
If your current score is at least 1, you may play the ith token face down, gaining tokens[i] power and losing 1 score.
Each token may be played at most once and in any order. You do not have to play all the tokens.

Return the largest possible score you can achieve after playing any number of tokens.

思路

从小到大sort tokens，取two pointers，每次花最小的token换score，否则用score换最多power。

答案

class Solution:
    def bagOfTokensScore(self, tokens, power):
        tokens.sort()
        l = 0
        r = len(tokens) - 1
        max_ans = score = 0
        
        while l <= r:
            if power >= tokens[l]:
                power -= tokens[l]
                score += 1
                max_ans = max(max_ans, score)
                l += 1
            elif score > 0:
                power += tokens[r]
                r -= 1
                score -= 1
            else:
                break
        
        return max_ans