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].

思路就是找到要merge的n个区间，然后把这几个区间删掉，再把新的区间插进去。

s初始化为-1，s+1就是插入的位置。s记录的是要merge的区间的前一个区间，所以插入位置就是s+1。

i就是要merge的区间的后一个区间。

如果merge错的话，可能会出现Output Limit Exceeded。

56ms。

 1 class Solution {
 2 public:
 3     vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
 4         int s = -1, i = 0;
 5         for (; i < intervals.size(); ++i) {
 6             if (newInterval.end < intervals[i].start) {
 7                 break;
 8             }
 9             if (newInterval.start > intervals[i].end) {
10                 s = i;
11                 continue;
12             }
13             
14             if (intervals[i].start < newInterval.start) newInterval.start = intervals[i].start;
15             if (intervals[i].end > newInterval.end) newInterval.end = intervals[i].end;
16         }
17         
18         if (i - s >= 2) intervals.erase(intervals.begin() + s + 1, intervals.begin() + i);
19         intervals.insert(intervals.begin() + s + 1, newInterval);
20         return intervals;
21     }
22 };

第三次写，用一个新的vector复制，因为vector的删除也是复制，这样还高效一点。

1. 不可能有交集的区间直接复制。

2. 有交集的区间求最终的交集。

3. 剩下的又直接复制。

这样可以不用对空输入进行特殊处理。

 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 class Solution {
11 public:
12     vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
13         vector<Interval> ans;
14         
15         int i = 0;
16         for (; i < intervals.size(); ++i) {
17             if (intervals[i].end < newInterval.start) {
18                 ans.push_back(intervals[i]);
19             } else if (intervals[i].start > newInterval.end) {
20                 break;
21             } else {
22                 if (intervals[i].end > newInterval.end) newInterval.end = intervals[i].end;
23                 if (intervals[i].start < newInterval.start) newInterval.start = intervals[i].start;
24             }
25         }
26         ans.push_back(newInterval);
27         for (;i < intervals.size(); ++i) {
28             ans.push_back(intervals[i]);
29         }
30         
31         return ans;
32     }
33 };