LeetCode 筆記 :(56) Merge Intervals

tags: Leetcode

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.

Leetcode Solution

官方的答案提供兩種不同的想法，因為Solution 2比較簡潔，所以我就簡單說明一下，先將list 針對地一個數進行排列。去比對每個list的end 是否小於下一個的start。如果是代表已經是不同區間。如果不是就進行合併。

Solution 2

class Solution:
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:

        intervals.sort(key=lambda x: x[0])

        merged = []
        for interval in intervals:
            # if the list of merged intervals is empty or if the current
            # interval does not overlap with the previous, simply append it.
            if not merged or merged[-1][1] < interval[0]:
                merged.append(interval)
            else:
            # otherwise, there is overlap, so we merge the current and previous
            # intervals.
                merged[-1][1] = max(merged[-1][1], interval[1])

        return merged