Leetcode 57. Insert Interval

最新推荐文章于 2025-08-21 10:38:03 发布

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最新推荐文章于 2025-08-21 10:38:03 发布

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原文链接：http://www.cnblogs.com/Deribs4/p/5735405.html

本文介绍了一个高效的算法来解决区间插入的问题。该算法适用于已排序的非重叠区间集合，并能够正确处理新插入区间的合并情况。通过寻找新区间的左右边界，并进行适当的区间合并，可以有效地更新整个区间列表。

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57. Insert Interval

Total Accepted: 64309
Total Submissions: 260619
Difficulty: Hard

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：区间划分。先找到新区间的左边界，再找到新区间的右边界。画图会更加清晰。

左边界：不满足intervals[i].end<newInterval.start意味着intervals[i-1].end<newInterval.start<=intervals[i].end。

右边界：满足intervals[j].start<newInterval.end时，不断合并区间，将intervals[j]和newInterval合并为新的newInterval，再尝试和intervals[j+1]合并，直到不能合并。因为intervals[j]和intervals[j+1]是区间独立的，所以intervals[j]被合并不会影响intervals[j+1]。

代码：

 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> insert(vector<Interval>& intervals, Interval newInterval) {
13         int i,j;
14         vector<Interval> res;
15         for(i=0;i<intervals.size()&&intervals[i].end<newInterval.start;i++){//i
16             res.push_back(intervals[i]);
17         }
18         for(j=i;j<intervals.size()&&intervals[j].start<=newInterval.end;j++){
19             newInterval=Interval(min(intervals[j].start,newInterval.start),max(intervals[j].end,newInterval.end));
20         }
21         res.push_back(newInterval);
22         for(;j<intervals.size();j++){
23             res.push_back(intervals[j]);
24         }
25         return res;
26     }
27 };