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].

 

Analyse: First sort, then merge. 

Notice the static function!

Runtime: 592ms

 1 /**
 2  * Definition for an interval.
 3  * struct 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 class Solution {
11 public:
12     static bool compare(Interval &interval1, Interval &interval2){
13         return interval1.start < interval2.start;
14     }
15     vector<Interval> merge(vector<Interval>& intervals) {
16         vector<Interval> result;
17         if(intervals.size() == 0) return result;
18         if(intervals.size() == 1) return intervals;
19         
20         sort(intervals.begin(), intervals.end(), Solution::compare);
21         Interval inter(intervals[0].start, intervals[0].end);
22         for(int i = 1; i < intervals.size(); i++){
23             if(inter.end < intervals[i].start){
24                 result.push_back(inter);
25                 inter.start = intervals[i].start;
26                 inter.end = intervals[i].end;
27             }
28             else{
29                 inter.start = min(inter.start, intervals[i].start);
30                 inter.end = max(inter.end, intervals[i].end);
31             }
32         }
33         result.push_back(inter);
34         return result;
35     }
36 };

 

 

Runtime: 596ms.

 1 /**
 2  * Definition for an interval.
 3  * struct 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 class Solution {
11 public:
12     vector<Interval> merge(vector<Interval>& intervals) {
13         vector<Interval> result;
14         if(intervals.empty()) return result;
15         
16         sort(intervals.begin(), intervals.end(), comp);
17         result.push_back(intervals[0]);
18         if(intervals.size() == 1) return result;
19         
20         for(int i = 1; i < intervals.size(); i++){
21             if(intervals[i].start > result[result.size() - 1].end)
22                 result.push_back(intervals[i]);
23             else{
24                 intervals[i].start = min(intervals[i].start, result[result.size() - 1].start);
25                 intervals[i].end = max(intervals[i].end, result[result.size() - 1].end);
26                 result.pop_back();
27                 result.push_back(intervals[i]);
28             }
29         }
30         return result;
31     }
32     
33     static bool comp(Interval int1, Interval int2){
34         return int1.start < int2.start;
35     }
36 };

 

posted @ 2015-08-10 18:55  amazingzoe  阅读(120)  评论(0)  编辑  收藏  举报