[leetcode] 435. Non-overlapping Intervals

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 <= 105
• intervals[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