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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.

Note that intervals which only touch at a point are non-overlapping. For example, [1, 2] and [2, 3] are 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
Related Topics
  • 贪心
  • 数组
  • 动态规划
  • 排序

  • 👍 1219
  • 👎 0

    • 思路 ​

    贪心算法，需要找到最满足策略的条件（最小，最大，最多，最少） 本地的约束条件是，找到最小的结束点，即可在同等情况下容纳更多元素

    解法 ​

    py
    # leetcode submit region begin(Prohibit modification and deletion)
from collections import defaultdict


class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        intervals.sort(key = lambda x: x[1])

        last = float('-inf')
        res = 0
        for start, end in intervals:
            if start >= last:
                last = end
            else:
                res += 1
        return res






# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度 O(nlogn)
    • 空间复杂度 O(1)