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[LeetCode] 56. Merge Intervals_Medium Tag: sort

2021-07-29 06:43  Johnson_强生仔仔  阅读(26)  评论(0编辑  收藏  举报

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