leetcode merge-intervals-优快云博客

本文链接：https://blog.youkuaiyun.com/pynash123/article/details/84572354

本文介绍了一种合并重叠区间的算法实现，通过排序和遍历的方式有效地合并了所有重叠区间。提供了完整的C++代码示例，并展示了如何通过比较区间的起始位置来判断是否进行合并。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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].

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
bool cmp(Interval I1, Interval I2)
{
    return I1.start<I2.start;
}
 
class Solution {
public:
    vector<Interval> merge(vector<Interval> &intervals) 
    {
        int cur = 0;
        vector<Interval> res;
        if (intervals.empty()) 
            return res;
        sort(intervals.begin(), intervals.end(), cmp);
        Interval tmp(intervals[0].start, intervals[0].end);
        for (int i=0;i<intervals.size()-1;i++)
        {
            if(tmp.end>=intervals[i+1].start)//一定要用tmp比较
            {
                tmp.start = min(tmp.start, intervals[i+1].start);//一定要用tmp比较
                tmp.end = max(tmp.end, intervals[i+1].end);//一定要用tmp比较
            }
            else
            {
                res.push_back(tmp);
                tmp.start = intervals[i+1].start;
                tmp.end = intervals[i+1].end;
            }
        }
        res.push_back(tmp);
        return res;
    }
};

测试

#include<algorithm>
#include<iostream>
#include<vector>
#include<limits.h>
using namespace std;



struct Interval {
    int start;
    int end;
    Interval() : start(0), end(0) {}
    Interval(int s, int e) : start(s), end(e) {}
};
 
bool cmp(Interval I1, Interval I2)
{
    return I1.start<I2.start;
}
 
class Solution {
public:
    vector<Interval> merge(vector<Interval> &intervals) 
    {
        int cur = 0;
        vector<Interval> res;
        if (intervals.empty()) 
            return res;
        sort(intervals.begin(), intervals.end(), cmp);
        Interval tmp(intervals[0].start, intervals[0].end);
        for (int i=0;i<intervals.size()-1;i++)
        {
            if(tmp.end>=intervals[i+1].start)
            {
                tmp.start = min(tmp.start, intervals[i+1].start);
                tmp.end = max(tmp.end, intervals[i+1].end);
            }
            else
            {
                res.push_back(tmp);
                tmp.start = intervals[i+1].start;
                tmp.end = intervals[i+1].end;
            }
        }
        res.push_back(tmp);
        return res;
    }
};

int main()
{
    Solution s;
    vector<Interval> intervals;
    Interval tmp1(0,2);
    intervals.push_back(tmp1);
    Interval tmp2(1,4);
    intervals.push_back(tmp2);
    Interval tmp3(3,5);
    intervals.push_back(tmp3);
   // Interval tmp4(18,19);
    //intervals.push_back(tmp4);
    
    vector<Interval>res = s.merge(intervals);
    
    for (int i=0;i<res.size();i++)
    {
        cout<<"start :"<<res[i].start<<" next :"<<res[i].end<<endl;
    }
}