[Leetcode 84] 56 Merge Intervals

合并区间问题详解

最新推荐文章于 2025-09-09 22:06:54 发布

转载最新推荐文章于 2025-09-09 22:06:54 发布 · 51 阅读

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原文链接：http://www.cnblogs.com/freeneng/p/3210450.html

文章标签：

#c/c++

本文详细介绍了如何解决合并区间的问题，并提供了一种有效的算法实现。首先对区间进行排序，然后遍历每个区间，若当前区间的开始时间在前一区间的结束时间内，则合并这两个区间；否则将前一区间加入结果集。此方法提高了处理效率。

Problem:

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].

Analysis:

First sort the vector according to the start time of each interval. Then scan through the vector, if the current interval's start time is between the former interval's strart and end time, we need to merge them. The newly merged interval's start is the former interval's start, but the end should be the max{current.end, former.end}.

To speed up this process, use an extra result vector to store the final result rather than using vector.erase method on original method.

Code:

 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  
11 bool cmpA(const Interval& a, const Interval& b)
12 {
13     return a.start < b.start;
14 }
15 
16 class Solution {
17 public:
18     vector<Interval> merge(vector<Interval> &intervals) {
19         // Start typing your C/C++ solution below
20         // DO NOT write int main() 
21         vector<Interval> res;
22         sort(intervals.begin(), intervals.end(), cmpA);
23 
24         Interval tmp = intervals[0];
25         for (int i=1; i<intervals.size(); i++) {
26             if (tmp.start <= intervals[i].start &&
27                 tmp.end >= intervals[i].start) {
28                     tmp.end = max(intervals[i].end, tmp.end);
29             }
30             else {
31                 res.push_back(tmp);
32                 tmp = intervals[i];
33             }
34         }
35 
36         return res;
37     }
38     
39     int max(int a, int b) {
40         return (a>b)? a : b;
41     }
42 };