20170401-leetcode-435-Non-overlapping Intervals
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1.Description
Given a collection of intervals, merge all overlapping intervals.
For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18]. Given a collection of intervals, merge all overlapping intervals.
For example,
Given [1,3],[2,6],[8,10],[15,18]
,
return [1,6],[8,10],[15,18]
.
https://leetcode.com/problems/merge-intervals/#/description
解读
给定一些间隔(interval),合并重复的区间
2.Solution
把其中的.end和.start分别换成[-1],[0]可以在自己的IDE上进行调试
有两种思路:1)按照end进行排序;2)按照start进行排序
2.1按照end进行排序
# Definition for an interval.# class Interval(object):# def __init__(self, s=0, e=0):# self.start = s# self.end = eclass Solution(object): def merge(self, intervals): """ :type intervals: List[Interval] :rtype: List[Interval] """ end = float('-inf') result = [] for interval in sorted(intervals, key=lambda interval: interval.end): #如果当前区间的左界大于上次的end值,那么判断该区间的左界跟result中栈顶间隔的左界大小 if interval.start <= end: #如果小于,那说明当前间隔包含了result的栈顶间隔,把栈顶间隔弹出, while result[-1].start>interval.start: result.pop() if result==[] or result[-1].end<interval.start: result.append(interval) break result[-1].end = interval.end else: result.append(interval) # 如果没有重叠,那么就把当前的interval添加到结果中 end = interval.end return result
2.2按照strat进行排序
class Solution(object): def merge(self, intervals): result = [] for interval in sorted(intervals, key=lambda interval: interval.start): if result and interval.start <= result[-1].end:#说明有重叠的部分 result[-1].end = max(result[-1].end, interval.end)#修改栈顶元素的右界值 else: result += interval, return result
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