[LeetCode]Non-overlapping Intervals

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:

给定一组区间,计算最少需要移除多少区间,才能使剩余区间两两互不相交。

注意:

  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:

贪心算法(Greedy Algorithm)

优先按照终点,然后按照起点从小到大排序

利用变量end记录当前的区间终点,end初始化为负无穷

遍历排序后的区间,若当前区间的起点≥end,则更新end为当前区间的终点,并将计数器ans+1

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

贪心算法的正确性:

假设输入区间包含m个不同的终止时间,据此可将它们划分为m个组(G1, G2, ..., Gm)

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

每一个分组内至多容纳一个区间,并且放弃检查其中任何一个区间并不会让最终答案更优。

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

 

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