[Leetcode/파이썬] 57. Insert Interval

Insert Interval

---

You are given an array of non-overlapping intervals intervals where intervals[i] = [starti, endi] represent the start and the end of the ith interval and intervals is sorted in ascending order by starti. You are also given an interval newInterval = [start, end] that represents the start and end of another interval.

Insert newInterval into intervals such that intervals is still sorted in ascending order by starti and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary).

Return intervals after the insertion.

Note that you don't need to modify intervals in-place. You can make a new array and return it.

 

Example 1:

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

Example 2:

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].

 

Constraints:

• 0 <= intervals.length <= 104
• intervals[i].length == 2
• 0 <= starti <= endi <= 105
• intervals is sorted by starti in ascending order.
• newInterval.length == 2
• 0 <= start <= end <= 105

 

class Solution:
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        res = []

        for interval in intervals :
            if newInterval[1] < interval[0] :
                res.append(newInterval)
                newInterval = interval
            elif newInterval[0] > interval[1] :
                res.append(interval)
            else :
                newInterval[0] = min(newInterval[0], interval[0])
                newInterval[1] = max(newInterval[1], interval[1])

        res.append(newInterval)
        return res

class Solution:
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        res = []

        intervals.append(newInterval)
        intervals.sort()

        res.append(intervals[0])

        for i in range(1, len(intervals)) :
            if res[-1][1] >= intervals[i][0] :
                res[-1][1] = max(res[-1][1], intervals[i][1])
            else :
                res.append(intervals[i])
        
        return res

두 풀이의 메커니즘은 동일하다. 범위를 하나 정하고 범위에 해당하면 해당 범위와 비교해서 최소값과 최대값으로 범위를 늘리거나 새로운 범위로 갱신한다. 만약 범위에 속하지 않는다면 범위를 res에 넣고, 새로운 범위로 다시 저장한다. 이는 intervals[i][0]으로 정렬되어 있기 때문에 이렇게 한 것이다. 만약 문제에 저런 조건이 주어지지 않았다면, intervals.sort()를 먼저 한 후에 진행하는 것이 좋겠다.