每日一题-区间合并

Merging Intervals

sort(a.begin(), a.end());
vector<pair<int, int> > res;
for (int i = 0; i < n; ++i) {
    int l = 0, r = res.size() - 1;
    
    while (l < r) {
        int mid = (l + r) >> 1;
        if (res[mid].second >= a[i].first) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    
    if (l < (int)res.size() and res[l].second >= a[i].first) {
        res[l].first = min(res[l].first, a[i].first);
        res[l].second = max(res[l].second, a[i].second);
    } else {
        res.push_back(a[i]);
    }
}
cout << res.size() << '\n';

Description

Given several intervals, u are asked the number of intervals if all intervals are merged. We can merge two intervals if they have overlapping parts.

Greedy & binary_search

1.First sort the intervals by left endpoint first, right endpoint second.
2.Add intervals into a container one by one, every time check if there exist an interval overlapping the one now adding, if it is, modify the interval into a bigger one, else simply add it. I use binary search to check it.