[leetcode] 1288. Remove Covered Intervals

[leetcode] 1288. Remove Covered Intervals

Description

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: 2Explanation: Interval [3,6] is covered by [2,8], therefore it is removed.

Example 2:

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

Example 3:

Input: intervals = [[0,10],[5,12]]Output: 2

Example 4:

Input: intervals = [[3,10],[4,10],[5,11]]Output: 2

Example 5:

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

Constraints:

1 <= intervals.length <= 1000intervals[i].length == 20 <= intervals[i][0] < intervals[i][1] <= 10^5All the intervals are unique.

分析

题目的意思是：删除区间数组里面覆盖的子区间，这道题类似合并区间的题目。

我借鉴了一下答案的思路，首先按照区间的开始端进行升序排序，如果区间是[start,end]的话，按照start先排序，然后按照end降序排序。排序完成后，开始遍历，只需要比较end就能够计算出需要保留的区间数了。

为啥这样排序，可以自行模拟一下，总的思路是按照区间起点进行排序，如果起点相同，先把终点大的保留，因为终点大的能够覆盖后面终点小的区间哈。

代码

class Solution: def removeCoveredIntervals(self, intervals: List[List[int]]) -> int: intervals.sort(key=lambda x:(x[0],-x[1])) res=0 prev_end=0 for _,end in intervals: if(end>prev_end): res+=1 prev_end=end return res

参考文献

​​删除被覆盖区间​​

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