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: 2
Explanation: 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 <= 1000
- intervals[i].length == 2
- 0 <= intervals[i][0] < intervals[i][1] <= 10^5
- All 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