Python贪心算法Greedy Algorithm解决案例小结

# 最好的选择是不要浪费饼干
class Solution:
    def findContentChildren(self, g: List[int], s: List[int]) -> int:
        # 先对胃口值和饼干尺寸排序
        g.sort()
        s.sort()
        g_l = len(g)
        g_index = 0
        s_l = len(s)
        s_index = 0
        # 计数
        count = 0
        # 终止条件：孩子数 和 饼干数是否在条件内
        while g_index < g_l and s_index < s_l:
            # 胃口小于饼干
            if g[g_index] <= s[s_index]:
                # 饼干被消耗
                count += 1
                g_index += 1
                s_index += 1
            # 胃口大于饼干
            else:
                # 寻求更多的饼干满足胃口
                s_index += 1
        return count

class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        # 判断是否为空
        if not intervals:
            return 0
        # 对end值进行升序排序
        intervals.sort(key = lambda x: x[1])
        # 维护一个最小值
        end_pos = intervals[0][1]
        # 只有单个区间时无重叠！！！因此定义为1
        count = 1
        # 终止条件
        for i in range(1, len(intervals)):
            # 判断是否连续
            if end_pos <= intervals[i][0]:
                count += 1
                end_pos = intervals[i][1]
        return len(intervals) - count

class Solution:
    def lemonadeChange(self, bills: List[int]) -> bool:
        five, ten, twenty = 0, 0, 0
        for bill in bills:
            if bill == 5:
                five += 1
            if bill == 10:
                # 是否可以找回
                if five <= 0:
                    return False
                # 收下 10 元
                ten += 1
                # 找回 5 元
                five -= 1
            if bill == 20:
                # 是否可以找回一张5元和一张10元
                if five > 0 and ten > 0:
                    five -= 1
                    ten -= 1
                    twenty += 1
                # 是否可以找回三张 5 元
                elif five >= 3:
                    five -= 3
                    twenty += 1
                else:
                    return False
        return True