56. Merge Intervals
Given a collection of intervals, merge all overlapping intervals.
For example,
Given [1,3],[2,6],[8,10],[15,18]
,
return [1,6],[8,10],[15,18]
.
代码如下:
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 @SuppressWarnings("unchecked") 12 public static List<Interval> partMerge(List<Interval> intervals){ 13 intervals.sort(new Comparator(){ 14 @Override 15 public int compare(Object o1, Object o2) { 16 Interval i1 = (Interval)o1; 17 Interval i2 = (Interval)o2; 18 if(i1.start!=i2.start) 19 return i1.start-i2.start; 20 else 21 return (Math.abs(i1.end-i1.start)-Math.abs(i2.end-i2.start)); 22 } 23 }); 24 25 List<Integer> removeList = new ArrayList<Integer>(); 26 for(int i = 0;i < intervals.size()-1; i++ ) 27 if(intervals.get(i).end>=intervals.get(i+1).start&&intervals.get(i).end<=intervals.get(i+1).end){ 28 intervals.get(i+1).start = intervals.get(i).start; 29 removeList.add(i); 30 }else if(intervals.get(i).end>=intervals.get(i+1).end){ 31 intervals.get(i+1).end = intervals.get(i).end; 32 intervals.get(i+1).start = intervals.get(i).start; 33 removeList.add(i); 34 } 35 for(int i = removeList.size()-1; i>=0 ; i--){ 36 intervals.remove(intervals.get(removeList.get(i))); 37 } 38 return intervals; 39 } 40 public List<Interval> merge(List<Interval> intervals) { 41 int presize = 0; 42 do{ 43 presize = intervals.size(); 44 intervals = partMerge(intervals); 45 }while(presize!=intervals.size()); 46 return intervals; 47 } 48 }
态度决定高度,细节决定成败,