[LeetCode]436. Find Right Interval 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/find-right-interval/description/ 题目描述: Given a set of intervals, for each of the interval i,…

Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the "right" of i. For any interval i, you ne…

Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the "right" of i. For any interval i, you ne…

Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the "right" of i. For any interval i, you ne…

Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the "right" of i. For any interval i, you ne…

给定一组区间,对于每一个区间 i,检查是否存在一个区间 j,它的起始点大于或等于区间 i 的终点,这可以称为 j 在 i 的“右侧”.对于任何区间,你需要存储的满足条件的区间 j 的最小索引,这意味着区间 j 有最小的起始点可以使其成为“右侧”区间.如果区间 j 不存在,则将区间 i 存储为 -1.最后,你需要输出一个值为存储的区间值的数组.注意:    你可以假设区间的终点总是大于它的起始点.    你可以假定这些区间都不具有相同的起始点.示例 1:输入: [ [1,2] ]输出: [-1]解…

题目如下: 解题思路:题目要求的是对于任意一个区间i,要找出一个区间j,使得j的起点最接近i的终点.既然这样,我们可以把所有区间的终点组成一个列表,并按大小排序,使用二分查找就可以快速找到j区间.注意要保存新的列表和输入的区间列表的元素映射关系,这样才能快速找到j区间在输入区间列表的索引. 代码如下: class Solution(object): def findRightInterval(self, intervals): """ :type intervals: Lis…

终于将LeetCode的免费题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开个题目汇总贴,并附上每道题目的解题连接,方便之后查阅吧~ 477 Total Hamming Distance 44.10% Meidum 475 Heaters  30.20% Easy 474 Ones and Zeroes  34.90% Meidum 473 Matchsticks to Square  31.80% Medium 472 Concatenated Words 29.20% Hard…

463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的边的数目即可.在查找重叠的边的数目的时候有一点小技巧,就是沿着其中两个方向就好,这种题目都有类似的规律,就是可以沿着上三角或者下三角形的方向来做.一刷一次ac,但是还没开始注意codestyle的问题,需要再刷一遍. class Solution { public: int islandPerime…

475. Heaters 思路:每趟循环查找离房子最近的热水器,计算距离,最后取最大距离 public int findRadius(int[] houses, int[] heaters) { Arrays.sort(houses); Arrays.sort(heaters); int j = 0; int res = 0; for(int i = 0; i < houses.length; i++){ //找离house[i]最近的heater while(j < heaters.leng…