56. Merge Intervals

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.

先用第一個值排序，再取第二個值比較多的部份。

# T: O(nlogn), S: O(n)
class Solution:
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
        
        # sort the intervals by their start value
        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

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