56 Merge Intervals-优快云博客

本文介绍了一种合并给定区间集合中所有重叠区间的算法。通过排序和遍历的方式，有效地将重叠区间进行合并，并提供了两种不同的实现方法。

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1 题目

Given a collection of intervals, merge all overlapping intervals.

Example 1:

Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:

Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

2 尝试解

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    void sort(vector<Interval>&  intervals,int start,int end){
        int left = start;
        int right = end;
        Interval pivot = intervals[left];
        if(left >= right) return;
        while(left<right){
            while(left < right && intervals[right].start > pivot.start){
                right --;
            }
            if(left<right) intervals[left] = intervals[right];
            while(left < right && intervals[left].start < pivot.start){
                left++;
            }
            if(left<right) intervals[right] = intervals[left];            
        }
        intervals[left] = pivot;
        sort(intervals,start,left-1);
        sort(intervals,left+1,end);
    }
    vector<Interval> merge(vector<Interval>& intervals) {
        if(intervals.size() < 2) return intervals;
        //sort vectors by the left bound
        for(int i = 0; i < intervals.size()-1;i++){
            for(int j = 0; j < intervals.size()-i-1;j++){
                if(intervals[j].start > intervals[j+1].start){
                    swap(intervals[j],intervals[j+1]);
                }
            }
        }
        
        for(int i = 0; i < intervals.size()-1; i++){
            if(intervals[i].end >= intervals[i+1].start){
                if(intervals[i].end < intervals[i+1].end)
                {intervals[i].end = intervals[i+1].end; }       //merge and remove
                intervals.erase(intervals.begin()+ i +1);
                i --;
            }
        }
        return intervals;
    }
};

3 标准解

vector<Interval> merge(vector<Interval>& ins) {
    if (ins.empty()) return vector<Interval>{};
    vector<Interval> res;
    sort(ins.begin(), ins.end(), [](Interval a, Interval b){return a.start < b.start;});
    res.push_back(ins[0]);
    for (int i = 1; i < ins.size(); i++) {
        if (res.back().end < ins[i].start) res.push_back(ins[i]);
        else
            res.back().end = max(res.back().end, ins[i].end);
    }
    return res;
}