classSolution{public:pair<int,int>findPos(vector<Interval> &intervals,intpos,intstart,intend){if(start>end)returnmake_pair(0, start);intmid = (start+end)/2;if(intervals[mid].start<=pos&&pos<=intervals[mid].end){
}classSolution {public://just reuse the solution of "Merge Intervals", quite straight forwardvector<Interval> insert(vector<Interval> &intervals, Interval newInterval) { intervals.push_back(newInterval);returnmerge(intervals); } vector<Interval> merge(vector<Interval> &intervals) { size_t n=inter...
return merge(intervals); } class LinkedNode { Interval val; LinkedNode next; public LinkedNode(Interval val, LinkedNode next) { this.val = val; this.next = next; } } public List<Interval> merge(List<Interval> intervals) { List<Interval> res = new ArrayList<Interval>(); if (intervals ...
* Interval(int s, int e) : start(s), end(e) {} * }; */classSolution{public://条件太多,每一个大小等号比较,每一个小下标就能让人栽跟斗。vector<Interval>insert(vector<Interval>&v,Interval nv){vector<Interval>res;intsta=nv.start,end=nv.end;inti=0;//注意:找到第一个起点的条件for(;...
high=intervals[i].end res.append([low,high])returnresif__name__=="__main__":intervals=Solution().insert([Interval(2,6),Interval(8,10),Interval(15,18)],Interval(13,16))forintervalinintervals:print(interval)
LeetCode-Insert Interval 对于一个排序的non-overlap的Interval集合。插入一个Interval。 遍历插入的情况。思路比较简单,代码如下: /** * Definition for an interval. * struct Interval { * int start; * int end; * Interval() : start(0), end(0) {}...
首先根据Interval的起点,我们将其排序,这样能合并的Interval就一定是相邻的了: [1,3] [5,6] [2,3] ---> [1,3] [2,3] [5,6] 然后我们就顺序遍历这个列表,并记录一个当前待合并的Interval,如果遍历到的Interval和当前待合并的Interval有重复部分,我们就将两个合并,如果没有重复部分,就将待合并的Inter...
Leetcode 57 Insert Interval sql 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 [2 ...
class Solution { public: vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) { int n = intervals.size(), leftEnd, rightEnd, l, r; vector<Interval> res; for (l = 0, r = n - 1; l <= r; ) { int mid = l + ((r - l) >> 1); if (intervals[mid]....
Insert Interval@LeetCode Insert Interval 这道题我今天重新看我以前提交的代码时,差点看吐了,巨复杂无比,先上代码,然后再分析为什么我当初会这样写。 javapublic class Solution { public List<Interval> insert(List<Interval> intervals, Interval newInterval) {...