題解

檢驗區間交疊，思路比較直觀，先排序再比較，看程式碼。 前面我的，後面的最精煉。

O(n^2)

class Solution {
public:
    static bool cmp(Interval &a,Interval &b){
        return a.start < b.start;
    }
    bool used[50000];
    vector<Interval> res;
    vector<Interval> merge(vector<Interval>& intervals) {
        sort
 
(intervals.begin(),intervals.end(),cmp);
        memset(used,false,sizeof(used));
        
        int st,ed;
        for(int i=0;i<intervals.size();i++){
            if( !used[i] ){
                st = intervals[i].start;
                ed = intervals[i].end;
                
                for
 
(int j=i+1;j<intervals.size();j++){
                     if( used[j] ) continue;
                     
                     if(intervals[j].start <= ed && intervals[j].end >ed){
                         used[j]= true;
                         ed = intervals[j].end;
                     }
 
 else if(intervals[j].start >= st && intervals[j].end <=ed){
                         used[j]= true;
                     }
                 }
                used[i]= true;
                res.push_back(Interval(st,ed) );
            }
        }
        return res;
    }
};

O(nlogn)

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;
}