LEETCODE-DAY35 - 技术栈

title: LEETCODE-DAY35

date: 2024-03-26 16:11:08
tags:

今日内容：860.柠檬水找零、406.根据身高重建队列、452. 用最少数量的箭引爆气球

T1

python 复制代码

class Solution:
    def lemonadeChange(self, bills: List[int]) -> bool:
        table={5:0,10:0,20:0}
        for i in range(len(bills)):
            if bills[i]==5:
                table[5]+=1
            elif bills[i]==10:
                if table[5]<=0:
                    return False
                else:
                    table[10]+=1
                    table[5]-=1
            elif bills[i]==20:
                if table[5]<=0 or table[10]<=0:
                    return False 
                else: 
                    table[20]+=1
                    table[10]-=1
                    table[5]-=1
        return True

输入

bills =

[5,5,10,20,5,5,5,5,5,5,5,5,5,10,5,5,20,5,20,5]

输出

false

预期结果

true

python 复制代码

class Solution:
    def lemonadeChange(self, bills: List[int]) -> bool:
        table={5:0,10:0,20:0}
        for i in range(len(bills)):
            if bills[i]==5:
                table[5]+=1
            elif bills[i]==10:
                if table[5]<=0:
                    return False
                else:
                    table[10]+=1
                    table[5]-=1
            elif bills[i]==20:
                if (5*table[5]+10*table[10]<15) or table[5]<=0:
                    return False 
                else: 
                    table[20]+=1
                    table[10]-=1
                    table[5]-=1
        return True

T2

本题主要难点在于如何将数组按照h从大到小，h相同时k从小到大的方式进行排序

python 复制代码

class Solution:
    def reconstructQueue(self, people: List[List[int]]) -> List[List[int]]:
        people.sort(key=lambda x:(-x[0], x[1]))
        queue=list()
        for i in range(len(people)):
            queue.insert(people[i][1],people[i])
        return queue

T3

python 复制代码

class Solution:
    def findMinArrowShots(self, points: List[List[int]]) -> int:
        points.sort()
        l=len(points)
        for i in range(l):
            if i>0 and points[i][0]<points[i-1][1]:
                points[i][1]=points[i-1][1]
                points.pop(i-1)

        return len(points)

IndexError: list index out of range

~~~~~~^^^

if i>0 and points[i][0]<points[i-1][1]:

队列长度减小了

python 复制代码

class Solution:
    def findMinArrowShots(self, points: List[List[int]]) -> int:
        points.sort()
        l=len(points)
        for i in range(l):
            if i>0 and i<len(points) and points[i][0]<points[i-1][1]:
                points[i][1]=points[i-1][1]
                points.pop(i-1)

        return len(points)

输入

points =

[[1,2],[2,3],[3,4],[4,5]]

输出

预期结果

python 复制代码

class Solution:
    def findMinArrowShots(self, points: List[List[int]]) -> int:
        points.sort()
        l=len(points)
        for i in range(l):
            if i>0 and i<len(points) and points[i][0]<=points[i-1][1]:
                points[i][1]=points[i-1][1]
                points.pop(i-1)

        return len(points)

points =

[[3,9],[7,12],[3,8],[6,8],[9,10],[2,9],[0,9],[3,9],[0,6],[2,8]]

输出

预期结果

问题在于points pop之后指标也会产生变化，所以需要定义一个变量来记录重叠次数

思路：没有重叠时需射l箭，每有一个重叠少射一箭

python 复制代码

class Solution:
    def findMinArrowShots(self, points: List[List[int]]) -> int:
        points.sort()
        l=len(points)
        j=0
        for i in range(l):
            if i>0 and points[i][0]<=points[i-1][1]:
                points[i][1]=min(points[i-1][1],points[i][1])
                j+=1

        return l-j

python 复制代码