1288. Remove Covered Intervals**

1288. Remove Covered Intervals**

https://leetcode.com/problems/remove-covered-intervals/

題目描述

Given a list of intervals, remove all intervals that are covered by another interval in the list. Interval [a,b) is covered by interval [c,d) if and only if c <= a and b <= d.

After doing so, return the number of remaining intervals.

Example 1:

Input: intervals = [[1,4],[3,6],[2,8]]
Output: 2
Explanation: Interval [3,6] is covered by [2,8], therefore it is removed.

Constraints:

• 1 <= intervals.length <= 1000
• 0 <= intervals[i][0] < intervals[i][1] <= 10^5
• intervals[i] != intervals[j] for all i != j

C++ 實現 1

使用最大堆. 對於區間 p = [x, y), 將 y 值較大的放到堆頂, 如果 y 值相等, 那麼將 x 值較小的放到堆頂. 使得堆頂的那個區間是覆蓋面最大的.

class Solution {
private:
    struct Comp {
        bool operator()(const vector<int> &p, const vector<int> &q) {
            return (p[1] < q[1]) || ((p[1] == q[1]) && (p[0] > q[0]));
        } 
    };
public:
    int removeCoveredIntervals(vector<vector<int>>& intervals) {
        priority_queue<vector<int>, vector<vector<int>>, Comp> q;
        for (auto &v : intervals) q.push(v);
        int res = 0;
        while (!q.empty()) {
            auto n = q.top();
            q.pop();
            res ++;
            // 之後只需要判斷堆頂的元素能覆蓋多少其他區間
            while (!q.empty() && n[0] <= q.top()[0]) q.pop();
        }
        return res;
    }
};

C++ 實現 2

排序的做法. 對於區間 p = [x, y), 將 x 較小的排在前面, 如果 x 相等, 則將 y 較大的排在前面. 這樣相當於固定了 x 的下限, 之後只需要用 ma 來記錄上一個覆蓋面最大的區間的 y 值, 對於當前訪問的區間 v, 它的 x 值肯定小於或等於上一個覆蓋面最大的區間的 x 值, 因此只需要判斷 v 的 y 值是否更大, 如果是的話, 說明沒有被上一個區間覆蓋.

class Solution {
public:
    int removeCoveredIntervals(vector<vector<int>>& intervals) {
        sort(intervals.begin(), intervals.end(), [](const vector<int> &v1, const vector<int> &v2)->bool {
            if (v1[0] == v2[0])
                return v1[1] > v2[1];
            return v1[0] < v2[0];
        });
        int ma = INT_MIN, res = 0;
        for (vector<int> v : intervals) {
            if (v[1] > ma) {
                res++;
                ma = v[1];
            }
        }
        return res;
    }
};