Description
You have an initial power of P, 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.
Example 1:
Input: tokens = [100], P = 50
Output: 0
Explanation: Playing the only token in the bag is impossible because you either have too little power or too little score.
Example 2:
Input: tokens = [100,200], P = 150
Output: 1
Explanation: Play the 0th token (100) face up, your power becomes 50 and score becomes 1.
There is no need to play the 1st token since you cannot play it face up to add to your score.
Example 3:
Input: tokens = [100,200,300,400], P = 200
Output: 2
Explanation: Play the tokens in this order to get a score of 2:
1. Play the 0th token (100) face up, your power becomes 100 and score becomes 1.
2. Play the 3rd token (400) face down, your power becomes 500 and score becomes 0.
3. Play the 1st token (200) face up, your power becomes 300 and score becomes 1.
4. Play the 2nd token (300) face up, your power becomes 0 and score becomes 2.
Constraints:
- 0 <= tokens.length <= 1000
- 0 <= tokens[i], P < 104
分析
题目的意思是：给定初始能量为 P，初始分数为 0，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。
- 先对tokens进行排序，然后从前面较小能量的值换分数，然后用分数换后面较大的能量，循环，贪心算法。可以用队列deque来实现，维护结果res就行了。
代码
class Solution:
def bagOfTokensScore(self, tokens: List[int], P: int) -> int:
tokens.sort()
deque=collections.deque(tokens)
res=0
score=0
while(deque and (P>=deque[0] or score)):
while(deque and P>=deque[0]):
P-=deque.popleft()
score+=1
res=max(res,score)
if(deque and score):
P+=deque.pop()
score-=1
return res