57. Insert Interval

发表于 | 分类于

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].

解法1：

自己写的版本比较繁琐，先用binary search来查找插入的位置，然后把interval插入，然后再merge interval。
C++

1

Java

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
public class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        List<Interval> res = new ArrayList<Interval>();
        if (intervals.size() == 0 && newInterval == null) {
            return res;
        } else if (intervals.size() == 0) {
            res.add(newInterval);
            return res;
        }
        int position = binarySearch(intervals, newInterval);
        // Insert into the intervals;
        if (position < intervals.size()) {
          intervals.add(position, newInterval);
        } else {
          intervals.add(newInterval);
        }
        // Merge Intervals using intervals
        Interval last = intervals.get(0);
        res.add(last);
        for (int i = 1; i < intervals.size(); i++) {
          Interval current = intervals.get(i);
          // check if we need to merge
          if (current.start <= last.end) {
            last.end = Math.max(last.end, current.end);
          } else {
            res.add(current);
            last = current;
          }
        }
        return res;
    }
    private int binarySearch(List<Interval> intervals, Interval newInterval) {
      // return the first interval that the start is larger than newInterval's start
      int start = 0, end = intervals.size() - 1;
      while (start + 1 < end) {
        int mid = start + (end - start) / 2;
        Interval temp = intervals.get(mid);
        if (temp.start <= newInterval.start) {
          start = mid;
        } else {
          end = mid;
        }
      }
      if (intervals.get(start).start >= newInterval.start) {
        return start;
      }
      if (intervals.get(end).start >= newInterval.start) {
        return end;
      }
      return intervals.size();
    }
}

解法2：

不需要用binary search来查找插入位置，而是在一边遍历的情况下用另一个指针pos来维护该插入的位置。这个思想在其他的题目中也见过。
Java

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
public class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        List<Interval> res = new ArrayList<Interval>();
        int pos = 0;    // find an insert position for newInterval
        for (Interval it : intervals) {
            if (it.end < newInterval.start) {
                res.add(it);
                pos++;
            } else if (newInterval.end < it.start) {
                res.add(it);
            } else {
                newInterval.start = Math.min(it.start, newInterval.start);
                newInterval.end = Math.max(it.end, newInterval.end);
            }
        }
        res.add(pos, newInterval);
        return res;
    }
}