1288. Remove Covered Intervals**

1288. Remove Covered Intervals**

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

题目描述

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++ 实现 1

使用最大堆. 对于区间 p = [x, y), 将 y 值较大的放到堆顶, 如果 y 值相等, 那么将 x 值较小的放到堆顶. 使得堆顶的那个区间是覆盖面最大的.

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 ++;
            // 之后只需要判断堆顶的元素能覆盖多少其他区间
            while (!q.empty() && n[0] <= q.top()[0]) q.pop();
        }
        return res;
    }
};

C++ 实现 2

排序的做法. 对于区间 p = [x, y), 将 x 较小的排在前面, 如果 x 相等, 则将 y 较大的排在前面. 这样相当于固定了 x 的下限, 之后只需要用 ma 来记录上一个覆盖面最大的区间的 y 值, 对于当前访问的区间 v, 它的 x 值肯定小于或等于上一个覆盖面最大的区间的 x 值, 因此只需要判断 v 的 y 值是否更大, 如果是的话, 说明没有被上一个区间覆盖.

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;
    }
};