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, losingtoken[i]
power, and gaining1
point. - If we have at least
1
point, we may play the token face down, gainingtoken[i]
power, and losing1
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:
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 = [100,200], P = 150 输出:1
示例 3:
输入:tokens = [100,200,300,400], 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.count 10 var p:Int = 0 11 var point:Int = 0 12 var ret:Int = 0 13 for i in 0...n 14 { 15 if i > 0 16 { 17 P += tokens[n-i] 18 point -= 1 19 } 20 while(p < n-i && P >= tokens[p]) 21 { 22 P -= tokens[p] 23 point += 1 24 p += 1 25 } 26 if p <= n-i 27 { 28 ret = max(ret, point) 29 } 30 } 31 return ret 32 } 33 }
原文地址:https://www.cnblogs.com/strengthen/p/10015319.html
时间: 2024-10-02 01:41:01