Insert intervals

最新推荐文章于 2025-08-16 15:48:41 发布

转载最新推荐文章于 2025-08-16 15:48:41 发布 · 98 阅读

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原文链接：http://www.cnblogs.com/tanghulu321/archive/2013/05/11/3072348.html

文章标签：

#c/c++

本文介绍了一种在已排序的非重叠区间中插入新区间并进行必要合并的算法。通过示例说明了如何处理新区间与已有区间的重叠情况，并提供了C++实现代码。

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题目：

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

思路：对于intervals从头到尾遍历，用new interval与相应的interval[i]对比，如果整个new interval都在interval[i]的左边，就把new interval存到result里，然后把剩下的所有interval都存起来即可，如果new interval在interval[i]的右边，则存起来interval[i]，然后继续遍历；如果new interval 和 interval[i]有overlap，则将两者进行合并，然后继续遍历。

代码：

 1 class Solution {
 2 public:
 3     vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
 4         // Start typing your C/C++ solution below
 5         // DO NOT write int main() function
 6         vector<Interval> result;
 7         
 8         for (int i = 0; i<intervals.size(); i++){
 9             
10             if (intervals[i].end < newInterval.start)
11                result.push_back(intervals[i]);
12             else if (intervals[i].start>newInterval.end){
13                 
14                 result.push_back(newInterval);
15                 while (i<intervals.size()){
16                     
17                     result.push_back(intervals[i]);
18                     i++;
19                 }
20             }
21             
22             else{
23                 
24                 newInterval.start = min(intervals[i].start, newInterval.start);
25                 newInterval.end = max(intervals[i].end, newInterval.end);
26                 
27             }
28         }
29         
30         if(0 == result.size() || result.back().end < newInterval.start)  
31             result.push_back(newInterval);  
32         
33         return result;
34     }
35 };