Leetcode-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].

Anaylsis:

Method 1: Sort the start points and end points respectively. For the ith point, if the (i-1)th end point is smaller than the ith start point, then the previous i-1 intervals are merged to one interval.

Method 2: Count the delta value at each time point. start: +1, end:-1. Then starting from the first time point, if the accumulative value reaches 0, then we get one interval.

Solution 1

 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> merge(List<Interval> intervals) {
12         List<Interval> res = new ArrayList<Interval>();
13         if (intervals.size()==0) return res;
14         if (intervals.size()==1){
15             res.add(intervals.get(0));
16             return res;
17         }
18         
19         
20         List<Integer> sList = new ArrayList<Integer>();
21         List<Integer> eList = new ArrayList<Integer>();
22         for (int i=0;i<intervals.size();i++){
23             sList.add(intervals.get(i).start);
24             eList.add(intervals.get(i).end);
25         }
26         Collections.sort(sList);
27         Collections.sort(eList);
28         
29         int p1=0;
30         while (p1<sList.size()){
31             int p2 = p1+1;
32             while (p2<sList.size() && sList.get(p2)<=eList.get(p2-1)) p2++;
33             Interval inter = new Interval(sList.get(p1),eList.get(p2-1));
34             res.add(inter);
35             p1 = p2;
36         }
37         
38         return res;
39     }
40 }

 

Solution 2:

 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> merge(List<Interval> intervals) {
12         Map<Integer,Integer> pMap = new HashMap<Integer,Integer>();
13         List<Integer> pList = new ArrayList<Integer>();
14 
15         //Count the delta at each point.
16         for (int i=0;i<intervals.size();i++){
17             Interval cur = intervals.get(i);
18             if (pMap.containsKey(cur.start)){
19                 pMap.put(cur.start,pMap.get(cur.start)+1);
20             } else {
21                 pList.add(cur.start);
22                 pMap.put(cur.start,1);
23             }
24             if (pMap.containsKey(cur.end)){
25                 pMap.put(cur.end,pMap.get(cur.end)-1);
26             } else {
27                 pList.add(cur.end);
28                 pMap.put(cur.end,-1);
29             }
30             }
31         
32         //Now, sort the time points.
33         Collections.sort(pList);
34         
35         //Now construct new interval list.
36         List<Interval> res = new ArrayList<Interval>();
37         int index1 = 0;
38         while (index1<pList.size()){
39             int val = pMap.get(pList.get(index1));
40             int index2 = index1;
41             while (val!=0){
42                 index2++;
43                 val += pMap.get(pList.get(index2));
44             }
45             Interval one = new Interval(pList.get(index1),pList.get(index2));
46             res.add(one);
47             index1 = index2+1;            
48         }
49         return res;
50     }
51 }

The most important thing here is to consider that the intervals are not sorted!.

posted @ 2014-11-14 11:05  LiBlog  阅读(132)  评论(0)  编辑  收藏  举报