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Leetcode 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 and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

 题目的意思:给定一个非重叠的区间,插入一个新的区间,如果区间与其他区间相交,则必须合并

注意题目有个假设,区间根据start时间进行了排序,所以自己不需要排序

struct Interval {    int start;    int end;    Interval() : start(0), end(0) {}    Interval(int s, int e) : start(s), end(e) {}};

解题思路:设给定的区间为intervals,插入的区间为newInterval,只需要将newInterval与intervals相交的区间合并即可,

如果newInterval与intervals没有相交的区间,则必须将newInterval插入到相应的位置

本题有三种情况

  (1)如果intervals[i].end < newInterval.start,说明intervals[i]与newInterval不相交,保留即可

  (2)如果intervals[i].start > newInterval.end, 说明intervals[i]与newInterval不相交,将newInterval插入即可,注意这时候其后面的intervals[i],都可以保留,这是由于本身区间是不想交的。

   (3)  如果intervals[i].start < newInterval.end,说明区间相交,合并区间,更新newInterval

class Solution {public:    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {        vector<Interval> res;        for(int i = 0 ; i < intervals.size(); ++ i){            if(intervals[i].end < newInterval.start) res.push_back(intervals[i]);            else if(intervals[i].start > newInterval.end) {                res.push_back(newInterval);                newInterval = intervals[i];            }else if(intervals[i].start <= newInterval.end ){                newInterval.start = min(newInterval.start, intervals[i].start);                newInterval.end =  max(newInterval.end, intervals[i].end);            }        }        res.push_back(newInterval);        return res;    }};