分析
初始化就是各个面额张数都是0
deposit就是对应加入面额
withdraw很关键
从大面额开始选
如果没有大面额了就continue了
然后看看大面额最多能选most，若超过张树，则设置为张树
然后amount 减去这么多张大面额的
同时used来记录当前用的
如果amount刚好为0了
就依次减掉原有张数（减掉used）
然后返回used
若不存在全部用完直接返回-1
ac code
class ATM:
def __init__(self):
self.c = [0, 0, 0, 0, 0]
#self.money = [20, 50, 100, 200, 500]
def deposit(self, banknotesCount: List[int]) -> None:
for i in range(5):
self.c[i] += banknotesCount[i]
#print(self.c)
def withdraw(self, amount: int) -> List[int]:
cur = self.c[:]
#print(now)
used = [0] * 5
money = [20, 50, 100, 200, 500]
for i in range(4, -1, -1):
if cur[i] == 0:
continue
most = amount // money[i]
if most >= cur[i]:
most = cur[i]
#print(most)
amount -= most * money[i]
#print(amount)
used[i] += most
if amount == 0:
for i in range(5):
self.c[i] -= used[i]
return used
return [-1]
总结
这道题不应该花这么多时间
贪心的知识不够牢固！
