leetcode1287

Posted 2021-03-23 AsenYang

 1 class Solution:
 2     def removeCoveredIntervals(self, intervals: ‘List[List[int]]‘) -> int:
 3         #intervals = sorted(intervals)
 4         n = len(intervals)
 5         dp = [0] * n
 6         for i in range(n):
 7             if dp[i] == 1:
 8                 continue
 9             else:
10                 for j in range(i+1,n):
11                     if dp[j] == 1:
12                         continue
13                     else:
14                         if intervals[i][0] <= intervals[j][0] and intervals[i][1] >=intervals[j][1]:
15                             dp[j] = 1
16                         elif intervals[i][0] >= intervals[j][0] and intervals[i][1] <=intervals[j][1]:
17                             dp[i] = 1
18         count = 0
19         for i in range(n):
20             if dp[i] == 0:
21                 count += 1
22         return count

1288. Remove Covered Intervals

并查集思想，将可以被合并的(范围小的)区间标记，最后没有标记的就是所求的个数。

 

另一种思路，先排序再比较，效率更高：

 1 class Solution:
 2     def removeCoveredIntervals(self, intervals: List[List[int]]) -> int:
 3         count = len(intervals)
 4         intervals.sort()
 5         MAX = intervals[0][1]
 6         for i,(s,e) in enumerate(intervals[1:]):
 7             if e > MAX:
 8                 MAX = e
 9             else:
10                 count -= 1
11         return count

参考：https://leetcode.com/problems/remove-covered-intervals/discuss/451532/Python3-O(n*logn)-solution-with-sort