标签:sort Medium Intervals List overlapping merge intervals ans
Given an array of intervals
where intervals[i] = [starti, endi]
, merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.
Example 1:
Input: intervals = [[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: intervals = [[1,4],[4,5]] Output: [[1,5]] Explanation: Intervals [1,4] and [4,5] are considered overlapping.
Constraints:
1 <= intervals.length <= 104
intervals[i].length == 2
0 <= starti <= endi <= 104
Ideas:
先sort intervals, 然后根据每个interval来判断跟ans里面最后一个是否重合, 如果重合, update ans[-1][1], 否则的话直接append进入到ans里面。
Code:
class Solution: def merge(self, intervals: List[List[int]]) -> List[List[int]]: intervals.sort(key = lambda x: tuple(x)) ans = [] for s, e in intervals: if not ans or s > ans[-1][1]: ans.append([s, e]) else: ans[-1][1] = max(ans[-1][1], e) return ans
标签:sort,Medium,Intervals,List,overlapping,merge,intervals,ans 来源: https://www.cnblogs.com/Johnsonxiong/p/15073422.html
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