[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].

先把要插入的区间放到最后，再排序，然后就跟上题一一模一样了。

 1 /**

 2  * Definition for an interval.

 3  * struct Interval {

 4  *     int start;

 5  *     int end;

 6  *     Interval() : start(0), end(0) {}

 7  *     Interval(int s, int e) : start(s), end(e) {}

 8  * };

 9  */

10  

11 bool cmp(const Interval &a, const Interval &b) {

12     return a.start < b.start;

13 }

14 

15 class Solution {

16 public:

17     vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {

18         int pos = 0, cnt = 0;

19         intervals.push_back(newInterval);

20         sort(intervals.begin(), intervals.end(), cmp);

21         for (int i = 1; i < intervals.size(); ++i) {

22             if (intervals[pos].end >= intervals[i].start) {

23                 ++cnt;

24                 if (intervals[pos].end < intervals[i].end) {

25                     intervals[pos].end = intervals[i].end;

26                 }

27             } else {

28                 ++pos;

29                 intervals[pos].start = intervals[i].start;

30                 intervals[pos].end = intervals[i].end;

31             }

32         }

33         intervals.resize(intervals.size()-cnt);

34         return intervals;

35     }

36 };