[Swift Weekly Contest 112]LeetCode948. 令牌放置 | Bag of Tokens

概述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 t

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 1point.

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:

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

你的初始能量为 P，初始分数为 0，只有一包令牌。

令牌的值为 token[i]，每个令牌最多只能使用一次，可能的两种使用方法如下：

如果你至少有 token[i] 点能量，可以将令牌置为正面朝上，失去 token[i] 点能量，并得到 1 分。 如果我们至少有 1 分，可以将令牌置为反面朝上，获得 token[i] 点能量，并失去 1 分。

在使用任意数量的令牌后，返回我们可以得到的最大分数。

示例 1：

输入：tokens = [100],P = 50输出：0

示例 2：

输入：tokens = [100,P = 150输出：1

示例 3：

输入：tokens = [100,P = 200输出：2

提示：

tokens.length <= 1000 0 <= tokens[i] < 10000 0 <= P < 10000 76ms

 1 class Solution { 2     func bagOfTokensscore(_ tokens: [Int],_ P: Int) -> Int { 3         var tokens = tokens.sorted(by:<) 4         var P = P 5         if tokens.count == 0 || P < tokens[0] 6         { 7             return 0 8         } 9         var n:Int = tokens.count10         var p:Int = 011         var point:Int = 012         var ret:Int = 013         for i in 0...n14         {15             if i > 016             {17                 P += tokens[n-i]18                 point -= 119             }20             while(p < n-i && P >= tokens[p])21             {22                 P -= tokens[p]23                 point += 124                 p += 125             }26             if p <= n-i27             {28                 ret = max(ret,point)29             }30         }31         return ret32     }33 }

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