Given a non-overlapping interval list which is sorted by start point.
Insert a new interval into it, make sure the list is still in order and non-overlapping (merge intervals if necessary).
Insert [2, 5] into [[1,2], [5,9]], we get [[1,9]].
Insert [3, 4] into [[1,2], [5,9]], we get [[1,2], [3,4], [5,9]].
Algorithm:
Iterate over the original interval list one by one and keep an insertion position for the new interval.
Case 1: newInterval is after the current interval: If the current interval‘s end < newInterval‘s start, insert the current interval to the new list, increment the insertion position by 1.
Case 2: newInterval is before the current interval: Else if the current interval‘s start > newInterval‘s end, insert the current interval to the new list.
Case 3: newInterval overlaps with the current interval. Update the newInterval‘s start and end to cover both intervals.
Repeat the above steps until all intervals in the original list are processed.
Finally, insert the newInterval to the saved insertion position.
1 /** 2 * Definition of Interval: 3 * public classs Interval { 4 * int start, end; 5 * Interval(int start, int end) { 6 * this.start = start; 7 * this.end = end; 8 * } 9 */ 10 public class Solution { 11 public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) { 12 if (newInterval == null) { 13 return intervals; 14 } 15 ArrayList<Interval> results = new ArrayList<Interval>(); 16 if(intervals == null) { 17 results.add(newInterval); 18 return results; 19 } 20 int insertPos = 0; 21 for (Interval interval : intervals) { 22 if (interval.end < newInterval.start) { 23 results.add(interval); 24 insertPos++; 25 } else if (interval.start > newInterval.end) { 26 results.add(interval); 27 } else { 28 newInterval.start = Math.min(interval.start, newInterval.start); 29 newInterval.end = Math.max(interval.end, newInterval.end); 30 } 31 } 32 results.add(insertPos, newInterval); 33 return results; 34 } 35 }
Related Problems
Merge Intervals
原文：http://www.cnblogs.com/lz87/p/7169574.html