20170401-leetcode-435-Non-overlapping Intervals

2017年4月1日 17:29:07

1.Description

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.

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.

解读

给出一些的间隔，去掉其中的n个，使得这些间隔彼此之间不重复，求n的最小值

2.Solution

这是给定的函数，直接调用

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

对intervals进行排序，排序的关键词为每个间隔的右阈值，这样每次比较当前间隔的左阈值和上一个间隔的右阈值，如果当前的比较大的话，说明没有问题，不会重复，否则的话输出值+1

class Solution(object):
def eraseOverlapIntervals(self, intervals):
"""
:type intervals: List[Interval]
:rtype: int
"""
end, num = float('-inf'), 0
for item in sorted(intervals, key=lambda item: item.end):
if item.start >= end:
end = item.end
else:
num += 1
return num

mark：

end=float(‘-inf’)

for item in sorted(intervals,key=lambda item:item.end)