leetcode56. Merge Intervals

题目要求：

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].

解题思路：

1）将集合进行排序；

集合的排序建议使用Collections的sort方法实现，需要自己实现Comparator接口；

2）依次进行归并，当后者的start大于当前Interval的end时，将当前Interval加入到集合中，否则将当前Interval和后者Interval进行归并

注意：边界条件的判断

代码：

 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
11
12 public class Solution {
13     public List<Interval> merge(List<Interval> intervals) {
14         //处理特殊情况
15         if(intervals == null || intervals.size() <2 ){
16             return intervals;
17         }
18
19         //集合排序
20         ComparatorImpl com = new ComparatorImpl();
21         Collections.sort(intervals,com);
22
23         //归并
24         List<Interval> temp = new ArrayList<Interval>();
25         Interval val = intervals.get(0);
26         Interval next;
27         boolean flag = true;
28         for(int i=1; i<intervals.size();){
29             next = intervals.get(i);
30             flag = true;
31             if(next.start > val.end){
32                 temp.add(val);
33                 val = next;
34                 flag = false;
35             }else{
36                 //进行一次归并，i自增一次
37                 val.end = Math.max(val.end,next.end);
38                 i++;
39             }
40         }
41
42         //处理最后一个未归并的Interval
43         if(flag){
44             temp.add(val);
45         }
46
47         return temp;
48     }
49 }
50
51 //实现排序接口
52 class ComparatorImpl implements Comparator<Interval>
53 {
54     public int compare(Interval o1, Interval o2) {
55         // TODO Auto-generated method stub
56         return o1.start-o2.start;
57     }
58
59 }

转载于:https://www.cnblogs.com/bywallance/p/5757242.html

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