【LeetCode】435. Non-overlapping Intervals 解题报告（Python）
作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/
题目地址：https://leetcode.com/problems/non-overlapping-intervals/description/
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:
- 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.
题目大意
给出了一些区间，求最少需要移除多少个区间才能保证所有的区间不重叠。如果两个区间的起始点与终点重复，不认为是重叠。
解题方法
看到区间最值题一般想到排序+贪心。这个题和452. Minimum Number of Arrows to Burst Balloons挺相似。
这个题的做法也不难，这么思考：我们尽量移除那些覆盖最广的区间。
先对所有区间的起点进行排序，然后进行遍历，如果新的区间起点比老的区间终点小的话，说明有重叠，需要移除区间，移除哪个区间呢？当然是终点最靠后的终点。
我们使用last指针指向上个保留下来的节点，如果intervals[i].end < intervals[last].end，则代表新的区间终点更靠前，所以使用第i个节点代表last，也就是说移除了上面的那个last。然后统计移除了多少次区间即可。
时间复杂度是O(nlogn + n)，空间复杂度是O(1).
代码如下：
# 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
日期
2018 年 9 月 16 日 ———— 天朗气清，惠风和畅
