python C++ Greedy速解找零問題Leetcode 860 Lemonade Change

python C++ Greedy速解找零問題Leetcode 860  Lemonade Change

b=20時貪婪演算法保持檢查順序，先找十塊，不要改變檢查順序，否則不起作用

[python解請進]

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When b=20, the greedy algorithm maintains the checking order. Give back $10 first, and do not change the checking order, otherwise it will not work.

[codes on Leetcode]https://leetcode.com/problems/lemonade-change/solutions/5637599/greedy-1-loop-vs-recursion-40ms-beats-99-76/

[Leetcode程式]https://www.youtube.com/watch?v=dybiq3pg-Os&list=PLYRlUBnWnd5IdDHk2BjqXwesydU17z_xk

class Solution:
    def lemonadeChange(self, bills: List[int]) -> bool:
        cnt5, cnt10=0, 0
        for b in bills:
            if b==5: cnt5+=1
            elif b==10:
                if cnt5>0:
                    cnt5-=1
                    cnt10+=1
                else:
                    return False
            else:
                if cnt10>0 and cnt5>0:
                    cnt10-=1
                    cnt5-=1
                elif cnt5>2: cnt5-=3
                else: return False
        return True