57. Insert Interval

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:

1 2	Input: intervals = [[1,3],[6,9]], newInterval = [2,5] Output: [[1,5],[6,9]]

Example 2:

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Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].

Example 3:

1 2	Input: intervals = [], newInterval = [5,7] Output: [[5,7]]

Example 4:

1 2	Input: intervals = [[1,5]], newInterval = [2,3] Output: [[1,5]]

Example 5:

1 2	Input: intervals = [[1,5]], newInterval = [2,7] Output: [[1,7]]

先加進去再回到56的方法排序。

# T: O(nlogn), S: O(n)
class Solution:
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        intervals.append(newInterval)
        intervals.sort(key=lambda x: x[0])
        merged = []
        for interval in intervals:
            if not merged or merged[-1][1] < interval[0]:
                merged.append(interval)
            else:
                merged[-1][1] = max(merged[-1][1], interval[1])
        return merged

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