Non-overlapping Intervals

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问题来源

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

Note:

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

Example 1:

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:

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:

Input: [ [1,2], [2,3] ]

Output: 0

Explanation: You don’t need to remove any of the intervals since they’re already non-overlapping.

直接看解法（来源）：

def eraseOverlapIntervals(self, intervals):
    end = float('-inf')
    erased = 0
    for i in sorted(intervals, key=lambda i: i.end):
        if i.start >= end:
            end = i.end
        else:
            erased += 1
    return erased

解释一下，代码逻辑很简单，就是两点，第一根据interval的end来排序，第二，遍历排序后的列表，如果当前interval的起始点大于等于终点，那么替换更新目标终点，否则计数加1。

为什么这个方法可以呢，注意到首先的排序就把所有的interval都排序了，计数统计的是交叉的个数，这要去掉交叉的个数那么肯定剩下的就是不交叉的。