## 题目描述：

LeetCode 435. Non-overlapping Intervals

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:

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.

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.
```

## 题目大意：

1. 你可以假设区间的终点总是比起点大
2. 区间[1,2]和[2,3]首尾相接但并不相交

## 解题思路：

ans为可以两两互不相交的最大区间数，len(intervals) - ans即为答案

`|---G1---|---G2---|---G3---|---...---|---Gm---|`

## Python代码：

``````# Definition for an interval.
# class Interval(object):
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

class Solution(object):
def eraseOverlapIntervals(self, intervals):
"""
:type intervals: List[Interval]
:rtype: int
"""
invs = sorted((x.end, x.start) for x in intervals)
end = -0x7FFFFFFF
ans = 0
for e, s in invs:
if s >= end:
end = e
ans += 1
return len(intervals) - ans
``````

Pingbacks已关闭。

1. 赤壁的火神 发布于 2016年10月31日 09:21 #

题主能说一下为什么会想到按照终点排序么 自己做的时候下意识地按照起点排序 然后有的测试用例就过不去了 现在都不知道错在哪里了 谢谢

2. 在线疯狂 发布于 2016年11月2日 13:58 #

按照起点排序很容易举出反例，我在博文里添加了一些说明。

3. 赤壁的火神 发布于 2016年11月3日 04:30 #

谢谢！