题目

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

思路

使用res列表存储结果，排序intervals序列并遍历，当当前元素与res列表有覆盖时，更新res，否则在res中追加当前元素。

代码

python版本：

class Solution:
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
        intervals = sorted(intervals, key=lambda x: x[0])
        res = []
        for i in range(len(intervals)):
            if not res or intervals[i][0] > res[-1][1]:
                res.append(intervals[i])
            elif res[-1][1] < intervals[i][1]:
                res[-1][1] = intervals[i][1]
        return res