题目描述 给出一个区间的集合,请合并所有重叠的区间。 示例 1: 输入: intervals = [[1,3],[2,6],[8,10],[15,18]] 输出: [[1,6],[8,10],[15,18]] 解释: 区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6]. 示例 2: 输入: intervals = [[1,4],[4,5]] 输出: [[1,5]] 解释: 区间 [1,4]
读完本文,你可以去力扣拿下如下题目: 56.合并区间 ----------- 上篇文章用贪心算法解决了区间调度问题:给你很多区间,让你求其中的最大不重叠子集。 其实对于区间相关的问题,还有很多其他类型,本文就来讲讲区间合并问题(Merge Interval)。 LeetCode 第 56 题就是一道相关问题,题目很好理解:
... M1: time = O(nlogn), space = O(logn) for sorting class Solution { public boolean canAttendMeetings(int[][] intervals) { Arrays.sort(intervals, (a, b) -> a[0] - b[0]); for(int i = 0; i + 1 < intervals.length; i++) {
Description Given interval sets \(A\) and \(B\) in [0,n], which contain \(x\) intervals and \(y\) intervals, respectively, find the number of common subintervals whose length is \(m\). Note that there are \(m\) points (including the start poi
问题描述 给出一个区间的集合,请合并所有重叠的区间。 示例 1: 输入: intervals = [[1,3],[2,6],[8,10],[15,18]] 输出: [[1,6],[8,10],[15,18]] 解释: 区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6]. 示例 2: 输入: intervals = [[1,4],[4,5]] 输出: [[1,5]] 解释: 区间 [1,4]
Given a collection of intervals, merge all overlapping intervals. Example 1: Input: intervals = [[1,3],[2,6],[8,10],[15,18]] Output: [[1,6],[8,10],[15,18]] Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6]. Example 2: Input: in
删除重叠的区间的个数 package my; import java.util.Arrays; public class NoOverlapIntervals2 { int eraseOverlapIntervals(int[][] intervals){ if(intervals.length == 0){ return 0; } //将所有区间按照起始时间排序
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 collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping. Example 1: Input: [[1,2],[2,3],[3,4],[1,3]] Output: 1 Explanation: [1,3] can be removed and the rest of intervals
Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen so far as a list of disjoint intervals. For example, suppose the integers from the data stream are 1, 3, 7, 2, 6, ..., then the summary will be: [1, 1] [1,
链接:https://leetcode-cn.com/problems/merge-intervals/ 代码 class Solution { public: vector<vector<int>> merge(vector<vector<int>>& intervals) { vector<vector<int>> ans; if (intervals.empty()) retur
package _interview_question /** * Problem: Find the most frequent element in all intervals Given an unsorted list of start and end time ( a range basically), find any number within all the ranges that occurs in maximum number of intervals. Example: [[1,
LeetCode中,有很多关于一组interval的问题。大体可分为两类: 1.查看是否有区间重叠; 2.合并重叠区间; 3.插入新的区间; 4. 基于interval的其他问题 【 做题通用的关键步骤】: 1. 按照begin时间排序; 2. 判断两个相邻区间是否重叠: 【假设】 a. 给定的区间是半开区间 [begin, end); b.
// 数轴上有n个闭区间[ai,bi]。取尽量少的点,使得每个区间内都至少有一个点(不同区间内含的点可以是同一个)。/*IntervalsYou are given n closed, integer intervals [ai, bi] and n integers c1, ..., cn.Write a program that:reads the number of intervals, their end points an
题目: 给出一个无重叠的 ,按照区间起始端点排序的区间列表。 在列表中插入一个新的区间,你需要确保列表中的区间仍然有序且不重叠(如果有必要的话,可以合并区间)。 思路:开始使用纯逻辑的思路,使用了大量的判断,如程序2所示,可以通过题目的两个例子,但是在自创例子时出现了超时的提示。看了一
力扣第56题 合并区间 class Solution { public: vector<vector<int>> merge(vector<vector<int>>& intervals) { sort(intervals.begin(), intervals.end(), [](const vector<int> &pl1, const vector<int> &pl2)
试题地址:https://leetcode-cn.com/problems/merge-intervals/ 试题思路:1、只有一个区域时,直接返回2、将区间以下界进行排序3、利用3维数组进行dp,每次比较临近两个区间,注意最后一个区间4、区间比较时,需要考虑第一个区间上界是大于第二个区间的下界还是上界比如输入:[[1 3] [15 18]
保证每次操作都是局部最优的,并且最后得到的结果是全局最优的。 1. 分配饼干 455. Assign Cookies (Easy) class Solution { public int findContentChildren(int[] g, int[] s) { if(g.length == 0 || s.length ==0){ return 0; } Array
Q:给定一组不重叠的时间区间,在时间区间中插入一个新的时间区间(如果有重叠的话就合并区间)。 这些时间区间初始是根据它们的开始时间排序的。 示例1: 给定时间区间[1,3],[6,9],在这两个时间区间中插入时间区间[2,5],并将它与原有的时间区间合并,变成[1,5],[6,9]. 示例2: 给定时间区间[1,
题目 给出一个区间的集合,请合并所有重叠的区间。 示例1 输入: [[1,3],[2,6],[8,10],[15,18]] 输出: [[1,6],[8,10],[15,18]] 解释: 区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6]. 示例2 输入: [[1,4],[4,5]] 输出: [[1,5]] 解释: 区间 [1,4] 和 [4,5] 可被视为重叠区间。
/** 56.合并区间 * @author 作者 Your-Name: * @version 创建时间:2020年2月19日 上午9:20:24 * 给出一个区间的集合,请合并所有重叠的区间。 示例 1: 输入: [[1,3],[2,6],[8,10],[15,18]] 输出: [[1,6],[8,10],[15,18]] 解释: 区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6]. 示
class Solution { public: vector<vector<int>> merge(vector<vector<int>>& intervals) { vector<vector<int>> res; if(intervals.empty()) return res; sort(intervals.begin(),interval
题目: Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary). You may assume that the intervals were initially sorted according to their start times. Example 1: Input: intervals = [[1,3],[6,9]], newInterval =
题目: Given a collection of intervals, merge all overlapping intervals. Example 1: Input: [[1,3],[2,6],[8,10],[15,18]] Output: [[1,6],[8,10],[15,18]] Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6]. Example 2: Input: [[1,4],[4,