Shirlies
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题目:

给出若干闭合区间,合并所有重叠的部分。

样例:

给出的区间列表 => 合并后的区间列表:

[                     [
  [1, 3],               [1, 6],
  [2, 6],      =>       [8, 10],
  [8, 10],              [15, 18]
  [15, 18]            ]
]

答案:

考虑输入可能为空的情况;先排序再遍历即可。

代码:

 1 /**
 2  * Definition of Interval:
 3  * classs Interval {
 4  *     int start, end;
 5  *     Interval(int start, int end) {
 6  *         this->start = start;
 7  *         this->end = end;
 8  *     }
 9  */
10 
11 #include <algorithm>
12 
13 using std::sort;
14 
15 bool cmp(const Interval &a, const Interval &b)
16 {
17     return a.start < b.start;
18 }
19 
20 class Solution {
21 public:
22     /**
23      * @param intervals: interval list.
24      * @return: A new interval list.
25      */
26     vector<Interval> merge(vector<Interval> &intervals) {
27          int i,j;
28        vector<Interval> ans;
29         if(intervals.empty())
30         {
31             return ans;
32         }
33        sort(intervals.begin(),intervals.end(),cmp);
34       
35        ans.push_back(intervals[0]);
36        
37        bool join;
38        for(i = 1; i < intervals.size(); ++ i)
39        {
40            join = false;
41            for(j = 0; j < ans.size(); ++ j)
42            {
43                if(intervals[i].start <= ans[j].end && intervals[i].end >= ans[j].end)
44                {
45                    ans[j].end = intervals[i].end;
46                    join = true;
47                    break;
48                }
49                
50                if(intervals[i].end <= ans[j].end)
51                {
52                    join = true;
53                    break;
54                }
55            }
56            
57            if(!join)
58            {
59                ans.push_back(intervals[i]);
60            }
61        }
62        
63        return ans;
64     }
65 };
View Code

 

posted on 2016-02-25 17:46  Shirlies  阅读(1131)  评论(0编辑  收藏  举报