1. 原题链接 https://leetcode.com/problems/insert-interval/description/ 2. 题目要求 该题与上一题的区别在于,插入一个新的interval对象,将此对象与所给列表里的Interval对象进行合并. 3. 解题思路 首先遍历Interval对象列表,找到插入位置.将插入位置之前的列表中的Interval对象直接加入结果列表. 然后查找所要插入Interval对象的end属性在列表中的位置. 最后将列表中剩余的Interval对象全部加…

原题地址 遍历每个区间intervals[i]: 如果intervals[i]在newInterval的左边,且没有交集,把intervals[i]插入result 如果intervals[i]在newInterval的右边,且没有交集,如果newInterval还没插入,则将newInterval插入result,之后再插入intervals[i] 如果intervals[i]和newInterval有交集,则更新newInterval的start.end 代码: vector<Interva…

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:Given intervals [1,3],[6,9], insert and merge […

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…

Insert Interval  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: Given intervals [1,3],[6,9],…

Insert interval  题意简述:给定若干个数轴上的闭区间,保证互不重合且有序,要求插入一个新的区间,并返回新的区间集合,保证有序且互不重合. 只想到了一个线性的解法,所有区间端点,只要被其他区间覆盖,就是不合法的,把他们去掉后,就可以直接得到答案.设新区间为[left,right],那么,比left小的端点显然合法,比right大的端点显然也合法,left和right之间的端点显然不合法,然后判断Left和right是否合法即可.难度不大,代码有些细节需要注意. /** * Defi…

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 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]], newInter…

lc57 Insert Interval 仔细分析题目,发现我们只需要处理那些与插入interval重叠的interval即可,换句话说,那些end早于插入start以及start晚于插入end的interval都可以保留.我们只需要两个指针,i&j分别保存重叠interval中最早start和最晚end即可,然后将interval [i, j]插入即可 class Solution { public int[][] insert(int[][] intervals, int[] newInte…

阅读了几个博客, 决定写一个简易版本; 忙着做更多题, 没有时间多考虑优化代码, 所以, 就写一个试试运气. http://blog.csdn.net/kenden23/article/details/17264441 http://blog.csdn.net/worldwindjp/article/details/21612731 Two versions are compared in the following blog: http://simpleandstupid.com/2014/1…