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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]].
分析：
Create a new array list, insert the smaller interval in the new array and use a counter to keep track of the total number of smaller intervals. If we find an interval overlapping with the new one, we need to change the start and end.
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 11 class Solution { 12 /** 13 * Insert newInterval into intervals. 14 * @param intervals: Sorted interval list. 15 * @param newInterval: A new interval. 16 * @return: A new sorted interval list. 17 */ 18 19 public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) { 20 if (newInterval == null || intervals == null) { 21 return intervals; 22 } 23 24 ArrayList<Interval> results = new ArrayList<Interval>(); 25 int insertPos = 0; 26 27 for (Interval interval : intervals) { 28 if (interval.end < newInterval.start) { 29 results.add(interval); 30 insertPos++; 31 } else if (interval.start > newInterval.end) { 32 results.add(interval); 33 } else { 34 newInterval.start = Math.min(interval.start, newInterval.start); 35 newInterval.end = Math.max(interval.end, newInterval.end); 36 } 37 } 38 39 results.add(insertPos, newInterval); 40 return results; 41 42 } 43 }
Given a collection of intervals, merge all overlapping intervals.
Given intervals => merged intervals:
[ [
[1, 3], [1, 6],
[2, 6], => [8, 10],
[8, 10], [15, 18]
[15, 18] ]
]
Analyze:
Sort first, then merge intervals if they overlap.
1 /** 2 * Definition of Interval: 3 * public class Interval { 4 * int start, end; 5 * Interval(int start, int end) { 6 * this.start = start; 7 * this.end = end; 8 * } 9 */ 10 11 class Solution { 12 /** 13 * @param intervals, a collection of intervals 14 * @return: A new sorted interval list. 15 */ 16 public List<Interval> merge(List<Interval> list) { 17 18 if (list == null || list.size() <= 1) { 19 return list; 20 } 21 22 Collections.sort(list, new Comparator<Interval>(){ 23 public int compare(Interval a, Interval b) { 24 return a.start - b.start; 25 }}); 26 27 for (int i = 1; i < list.size(); i++) { 28 //no intersection 29 if (overlap(list.get(i), list.get(i - 1))) { 30 list.get(i - 1).start = Math.min(list.get(i).start, list.get(i - 1).start); 31 list.get(i - 1).end = Math.max(list.get(i).end, list.get(i - 1).end); 32 list.remove(i); 33 i--; 34 } 35 } 36 return list; 37 } 38 39 boolean overlap(Interval i1, Interval i2) { 40 return Math.max(i1.start, i2.start) <= Math.min(i1.end, i2.end); 41 } 42 }
Insert Interval & Merge Intervals
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原文地址：http://www.cnblogs.com/beiyeqingteng/p/5668863.html
