c_cpp 如果你有一个约会列表（每个约会有一个开始时间，一个结束时间和一个布尔值hasConflict），你将如何有效地g

Posted 2021-05-14

/*
If you had a list of appointments (each appointment has a begin time, an end time, and a boolean hasConflict), how would you efficiently go through them and set the hasConflict boolean for each. You cannot assume they are sorted in any way. Keep in mind that one appointment may be very long, etc.
*/

struct appointment {
    int begin, end;
    bool has_conflict;
}

bool is_overlap(appointment &a, appointment &b) {
    if(a.end < b.begin || a.begin > b.end) return false;
    return true;
}

// solution 1, brute force. compare every pair.

void update(vector<appointment>& va) {
    if(va.empty()) return;
    for(int i=0; i<va.size()-1; i++) {
        for(int j=i+1; j<va.size(); j++) {
            if(va[j].has_conflict == false && is_overlap(va[i], va[j]) == true)
                va[i].has_conflict = va[j].has_conflict = true;
        }
    }
}

void update_better(vector<appointment>& va) {
    if(va.empty()) return;
    sort(va.begin(), va.end(), [](appointment& a, appointment&b) { return a.begin < b.begin; } );
    for(int i=0; i<va.size()-1; i++) {
        for(int j=i+1; j<va.size(); j++) {
            if(is_overlap(va[i], va[j]) == false) break;
            else if(va[j].has_conflict == false) {
                va[i].has_conflict = va[j].has_conflict = true;
            }
        }
    }
}