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57 Insert Interval

1. Question

Given a set ofnon-overlappingintervals, 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].

2. Implementation

(1) Scan Line
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/**
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* Definition for an interval.
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* public class Interval {
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* int start;
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* int end;
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* Interval() { start = 0; end = 0; }
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* Interval(int s, int e) { start = s; end = e; }
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* }
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*/
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class Solution {
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public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
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List<Interval> res = new ArrayList<>();
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for (Interval interval : intervals) {
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if (newInterval == null || interval.end < newInterval.start) {
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res.add(interval);
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}
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else if (newInterval.end < interval.start) {
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res.add(newInterval);
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res.add(interval);
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newInterval = null;
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}
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else {
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newInterval.start = Math.min(interval.start, newInterval.start);
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newInterval.end = Math.max(interval.end, newInterval.end);
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}
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}
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if (newInterval != null) {
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res.add(newInterval);
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}
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return res;
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}
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}
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3. Time & Space Complexity

Scan Line: 时间复杂度O(n), 空间复杂度O(n)