# # 【LeetCode】435. Non-overlapping Intervals 解题报告（Python）

## # 题目描述：

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.
``````

## # 解题方法

``````# 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
"""
if not intervals: return 0
intervals.sort(key = lambda x : x.start)
last = 0
res = 0
for i in range(1, len(intervals)):
if intervals[last].end > intervals[i].start:
if intervals[i].end < intervals[last].end:
last = i
res += 1
else:
last = i
return res
``````

http://www.cnblogs.com/grandyang/p/6017505.html

## # 日期

2018 年 9 月 16 日 ———— 天朗气清，惠风和畅