[Leetcode]56. Merge Intervals

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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 considerred overlapping.

思路很简单，就是首先排序，这里为了练习快排，所以自己手写了以个排序。

针对intervals进行排序，然后进行类似于mapreduce中的reduce工作一样，进行归并，得到结果，具体看代码。

const int x=[](){
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    return 0;
}();
class Solution {
  
public:
    void swap(Interval &a, Interval &b){
        Interval temp=a;
        a=b;
        b=temp;
    }
    void mysqrt(vector<Interval>& intervals, int left, int right){
        if(left>=right) return;
        int i=left-1;
        int j=right;
        for(int k=left;k<right;k++){
            if(intervals[k].start<intervals[right].start){
                i++;
                swap(intervals[k],intervals[i]);
            }
        }
        i++;
        swap(intervals[i],intervals[right]);
        mysqrt(intervals,left,i-1);
        mysqrt(intervals,i+1,right);
        
    }
    vector<Interval> merge(vector<Interval>& intervals) {
        if(intervals.empty()) return intervals;
        mysqrt(intervals,0,intervals.size()-1);
        vector<Interval> ans;
        Interval cur=intervals[0];
        for(int i = 1; i < intervals.size(); i++){
            if(intervals[i].start <= cur.end){
                cur.end=intervals[i].end > cur.end? intervals[i].end : cur.end;
            }else{
                ans.push_back(cur);
                cur=intervals[i];
            }
        }
        ans.push_back(cur);
        return ans;
    }
};