标签：inter xpl val interval into ide 初始 amp col
Given a collection of intervals, merge all overlapping intervals.
Example 1:
Input: [[1,3],[2,6],[8,10],[15,18]] Output: [[1,6],[8,10],[15,18]] Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
Example 2:
Input: [[1,4],[4,5]] Output: [[1,5]] Explanation: Intervals [1,4] and [4,5] are considered overlapping.
思路：
对所有按键值排序，看下一个是和它合并还是单独存储
class Solution(object): def merge(self, intervals): ans = [] # 使用下标表示interval里的是按哪个排序 for interval in sorted(intervals, key=lambda x: x[0]): # 如果初始或者当前与上一个足够拉开差距，直接压入 if not ans or interval[0] > ans[-1][1]: ans.append(interval) # 否则扩展上一个ans的范围 else: ans[-1][1] = max(ans[-1][1], interval[1]) return ans solu=Solution() intervals=[[1,3],[2,6],[8,10],[15,18]] print(solu.merge(intervals))
LeetCode开心刷题四十二天——56. Merge Intervals
标签：inter xpl val interval into ide 初始 amp col
原文地址：https://www.cnblogs.com/Marigolci/p/11525461.html
