题目
Given an array of intervals intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
Example 1:
Input: intervals = [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping.
Example 2:
Input: intervals = [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.
Example 3:
Input: intervals = [[1,2],[2,3]]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
Constraints:
1 <= intervals.length <= 105intervals[i].length == 2-5 * 104 <= starti < endi <= 5 * 104
思路
在众多的intervals中，总是优先拿end比较小的interval，因为end越小，表明接下来所能够提供的容量越大。
将intervals序列按end排序，随后遍历序列，当遇到存在覆盖的interval时，则删除它，最后记录删除的interval的个数。
代码
python版本：
class Solution:
def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
end, cnt = -math.inf, 0
for x, y in sorted(intervals, key=lambda x: x[1]):
if x >= end:
end = y
else:
cnt += 1
return cnt
