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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].
这道题可以在Merge Intervals基础上完成，时间复杂度有点高。
1 /** 2 * Definition for an interval. 3 * public class 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 public class Solution { 11 public List<Interval> insert(List<Interval> intervals, Interval newInterval) { 12 int low = newInterval.start; 13 int high = newInterval.end; 14 ListIterator<Interval> iterator = intervals.listIterator(); 15 16 while(iterator.hasNext()){ 17 Interval interval = iterator.next(); 18 if(high < interval.start){ 19 iterator.previous(); 20 iterator.add(new Interval(low, high)); 21 return intervals; 22 } 23 if(low > interval.end) 24 continue; 25 else{ 26 low = Math.min(low, interval.start); 27 high = Math.max(high, interval.end); 28 iterator.remove(); 29 } 30 }//while 31 intervals.add(new Interval(low, high)); 32 33 return intervals; 34 } 35 }
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原文地址：http://www.cnblogs.com/luckygxf/p/4255217.html
