Leetcode 1288. Remove Covered Intervals (区间好题)

原创已于 2024-01-17 15:09:02 修改 · 452 阅读

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于 2023-12-16 16:57:09 首次发布

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文章讨论了如何在一个给定的区间数组中，移除被其他区间完全覆盖或部分覆盖的区间，通过两种不同的排序方法实现解决方案，最后返回剩余区间的数量。

Remove Covered Intervals
Medium
Given an array intervals where intervals[i] = [li, ri] represent the interval [li, ri), remove all intervals that are covered by another interval in the list.

The interval [a, b) is covered by the interval [c, d) if and only if c <= a and b <= d.

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.
Example 2:

Input: intervals = [[1,4],[2,3]]
Output: 1

Constraints:

1 <= intervals.length <= 1000
intervals[i].length == 2
0 <= li < ri <= 105
All the given intervals are unique.

解法1：先把left和right设为intervals[0]的边界，然后再分1) cover 2) overlap 3) 无重合 3种情况处理。

struct cmp {
    bool operator () (const vector<int> &a, const vector<int> &b) {
        if (a[0] <= b[0]) return true;
        if (a[0] == b[0]) return a[1] < b[1];
        return false;
    }
};

class Solution {
public:
    int removeCoveredIntervals(vector<vector<int>>& intervals) {
        int n = intervals.size();
        if (n == 0) return 0;
        sort(intervals.begin(), intervals.end(), cmp()); //sort的第3个参数要么是function()，要么是类或结构的实例！但priority_queue的第3个参数是类或结构的定义！
        //sort(intervals.begin(), intervals.end(), [](const auto &a, const auto &b){ return a[0] < b[0];}); //写成这样更简练
        int left = intervals[0][0] , right = intervals[0][1];
        int res = 0;
        for (int i = 1; i < n; i++) {
            if (left <= intervals[i][0] && right >= intervals[i][1]) { //case 1: cover
                res++;
            } else if (left < intervals[i][1] && right > intervals[i][0]) { //case 2: overlap
                left = min(left, intervals[i][0]);
                right = max(right, intervals[i][1]);
            } else { //case 3: no cover or overlap
                left = intervals[i][0];
                right = intervals[i][1];
            }
        }
        return n - res;
    }
};

二刷：


class Solution {
public:
    int removeCoveredIntervals(vector<vector<int>>& intervals) {
        int n = intervals.size();
        if (n == 0) return 0;
        sort(intervals.begin(), intervals.end(), [](vector<int>&a, vector<int>&b){
            if (a[0] == b[0]) return a[1] > b[1]; //如果两个区间起始位置相同，把大的区间排在前面
            return a[0] < b[0];
        });
        int count = n;
        vector<int> interval = intervals[0];
        for (int i = 1; i < n; i++) {
            if (interval[1] >= intervals[i][1]) {
                count--;
             } else if (interval[1] <= intervals[i][0]){
                interval = intervals[i];
            } else {
                interval[1] = intervals[i][1];
            }
        }
        return count;
    }
};