[LeetCode] 56. Merge Intervals_Medium Tag: sort

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