There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinates of s…

There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinates of s…

[LeetCode]452. Minimum Number of Arrows to Burst Balloons 解题报告(Python) 标签(空格分隔): LeetCode 题目地址:https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/description/ 题目描述: There are a number of spherical balloons spread in two-dimensio…

870. Advantage Shuffle 思路:A数组的最大值大于B的最大值,就拿这个A跟B比较:如果不大于,就拿最小值跟B比较 A可以改变顺序,但B的顺序不能改变,只能通过容器来获得由大到小的顺序,并且必须存储相应的index,因为最终需要将选择的A的数值存入与这个B相对应的index下 class Solution { public: vector<int> advantageCount(vector<int>& A, vector<int>&…

There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinates of s…

There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinates of s…

There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinates of s…

［抄题］: There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinate…

There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinates of s…

题目如下: 解题思路:本题可以采用贪心算法.首先把balloons数组按end从小到大排序,然后让第一个arrow的值等于第一个元素的end,依次遍历数组,如果arrow不在当前元素的start到end的区间,表示这个arrow不能刺破气球,arrow总数加一,然后令arrow继续等于当前这个元素的end,直到数组遍历完成. 代码如下: class Solution(object): def findMinArrowShots(self, points): """ :type…