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 = [2,5]
Output: [[1,5],[6,9]]

Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]]
 
, newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].

題意

給定一組以左邊界大小排列的互不重合的閉區間，求加入一個新區間後的集合（也需要從小到大排列）

題解

一開始做的十分復雜

 1 class Solution {
 2 public:
 3     vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
 
 4         if (intervals.size() < 1) {
 5             intervals.push_back(newInterval);
 6             return intervals;
 7         }
 8         int s = newInterval.start, e = newInterval.end, i;
 9         for (i = 0; i < intervals.size(); i++) {
10             if (intervals[i].start <= s && intervals[i].end >= s) {
 
11                 if (e <= intervals[i].end)
12                     return intervals;
13                 break;
14             }
15             else if (intervals[i].start > s) {
16                 if(intervals[i].start>e) {
17                     intervals.insert(intervals.begin()+i,newInterval);
18                     return intervals;
19                 }
20                 intervals[i].start = s;
21                 if (e <= intervals[i].end)
22                     return intervals;
23                 break;
24             }
25         }
26         int mark = i;
27         i++;
28         for (; i < intervals.size(); ) {
29             if (intervals[i].start > e) {
30                 intervals[mark].end = e;
31                 return intervals;
32             }
33             else if (intervals[i].end >= e) {
34                 intervals[mark].end = intervals[i].end;
35                 intervals.erase(intervals.begin() + i);
36                 return intervals;
37             }
38             intervals.erase(intervals.begin() + i);
39         }
40         if(mark>=intervals.size())
41             intervals.push_back(newInterval);
42         else
43             intervals[mark].end = e;
44         return intervals;
45     }
46 };

需要考慮很多條件，於是重寫

 1 class Solution {
 2 public:
 3     vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
 4         int i = 0, s = newInterval.start, e = newInterval.end;
 5         vector<Interval>ans;
 6         while (i < intervals.size() && intervals[i].end < s) {
 7             ans.push_back(intervals[i]);
 8             i++;
 9         }
10         int left = s, right = e;
11         while (i < intervals.size() && intervals[i].start <= e) {
12             left = min(left, intervals[i].start);
13             right = max(right, intervals[i].end);
14             i++;
15         }
16         ans.push_back(Interval(left, right));
17         while (i < intervals.size()) {
18             ans.push_back(intervals[i]);
19             i++;
20         }
21         return ans;
22     }
23 };

這個就快得多了……

19.2.9 [LeetCode 57] Insert Interval