LeetCode 435. Non-overlapping Intervals

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题目

Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Example 1:

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Input: [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.

Example 2:

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Input: [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.

Example 3:

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Input: [[1,2],[2,3]]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

Note:

  1. You may assume the interval’s end point is always bigger than its start point.
  2. Intervals like [1,2] and [2,3] have borders “touching” but they don’t overlap each other.

思路

Sort + Greedy.

代码

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class Solution(object):
    def eraseOverlapIntervals(self, intervals):
        """
        :type intervals: List[List[int]]
        :rtype: int
        """
        if not intervals: 
            return 0
        intervals.sort(key = lambda x : x[0])
        last = 0
        res = 0
        for i in range(1, len(intervals)):
            if intervals[last][1] > intervals[i][0]:
                if intervals[i][1] < intervals[last][1]:
                    last = i
                res += 1
            else:
                last = i
        return res