435. Non-overlapping Intervals
难度: 中等
刷题内容
原题连接
- https://leetcode.com/problems/non-overlapping-intervals
内容描述
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.
解题方案
思路 1
先按照start排序，然后每次如果下一个的start小于前一个的end的时候意味着我们需要删掉一个了，但是我们尽量留下end比较小的那个Interval，具体看代码会比较清晰
# 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 or len(intervals) == 0:
return 0
res = 0
intervals = sorted(intervals, key=lambda x:x.start)
cur_end = intervals[0].end
for i in range(1, len(intervals)):
if intervals[i].start < cur_end: #overlap
res += 1
cur_end = min(intervals[i].end, cur_end) ## 尽量留下end小的Interval
else:
cur_end = intervals[i].end
return res
思路 2
又发现有大佬比我牛p多了，代码更nice，跟646题比较像
首先按照end排序，我们可以知道的，一旦后面一个Interval的start比前一个的end小的话，这个时候我们就需要删除掉当前的这个Interval, 反之则前面的那些Interval已经成立了，我们只需要更新cur_end为当前Interval的end
class Solution(object):
def eraseOverlapIntervals(self, intervals):
"""
:type intervals: List[Interval]
:rtype: int
"""
intervals.sort(key = lambda x: x.end)
res, cur_end = 0, -float("inf")
for i in intervals:
if cur_end > i.start: res += 1
else: cur_end = i.end
return res
