这道题使用了构建了字典去查询的
            
            
              python
              
              
            
          
          class Solution:
    def lemonadeChange(self, bills: List[int]) -> bool:
        yours = {"5":0,"10":0,"20":0}
        for i in range(len(bills)):
            if bills[i] == 5:
                yours["5"] += 1
            elif bills[i] == 10:
                if yours["5"] != 0:
                    yours["5"] -= 1
                    yours["10"] += 1
                else:
                    return False
            else:
                if yours["5"] != 0 and yours["10"] != 0:
                    yours["20"] += 1
                    yours["5"] -= 1
                    yours["10"] -= 1
                elif yours["5"] >= 3:
                    yours["5"] -= 3
                    yours["20"] += 1
                else:
                    return False
        print(yours)
        return True
            
            
              python
              
              
            
          
          class Solution:
    def reconstructQueue(self, people: List[List[int]]) -> List[List[int]]:
        people.sort(key=lambda x: (-x[0], x[1]))
        que = []
        for p in people:
            que.insert(p[1], p)
        return que
        感觉这道题十分简单,但是有想不到,贪心有时就是这样
可能是因为贪心不能有一个固定的公式,所以总觉得做的很吃力,完全没有头绪,没有一个可以套的公式。
            
            
              python
              
              
            
          
          class Solution:
    def findMinArrowShots(self, points: List[List[int]]) -> int:
        points.sort(key = lambda x:(x[0],x[1]))
        i = 1
        result = 1
        for i in range(1, len(points)):
            if points[i][0] > points[i - 1][1]:
                result += 1     
            else:
                points[i][1] = min(points[i - 1][1], points[i][1]) # 更新重叠气球最小右边界
        return result