動機

當初想用處理interval的老套路，stack去做，一直gg 直到看解答才發現，原來這麼簡單

Problem

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: 1Explanation: [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: 2Explanation: 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: 0Explanation: 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

Sol

  • 如果有overlap就一定會刪一個
    • 刪end最大的那一個
class Solution:
    def eraseOverlapIntervals(self, ins: List[List[int]]) -> int:
        ins.sort(key=lambda x: x[0])
        
        ret = 0
        last = 0
        for i in range(1,len(ins)):
            if ins[i][0] < ins[last][1]:
                ret += 1
                if ins[i][1] < ins[last][1]:
                    last = i
            else:
                last = i
        return ret