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

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:

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

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