[LeetCode / Kotlin] Merge Intervals

문제

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] overlap, 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
  
  

나의 풀이

class Solution {
    fun merge(intervals: Array<IntArray>): Array<IntArray> {
        // 겹치는 범위가 있으면 통합시켜라
        // 겹친다 -> start와 end에 대한 비교
        // 비교는 어떻게? -> end 오름차순으로 정렬 후, 반복문을 돈다. 이때, 겹치면 커서end값 갱신, 아니면 커서 start, 커서 end 갱신
        val sortedIntervals = intervals.sortedBy { it[0] }
        val ans = mutableListOf<IntArray>()

        var (cursorS, cursorE) = sortedIntervals.first()
        for (i in 1 until sortedIntervals.size) {
            val (start, end) = sortedIntervals[i]

            if (cursorE >= start) {
                cursorE = maxOf(cursorE, end)
            } else {
                ans.add(intArrayOf(cursorS, cursorE))
                cursorS = start
                cursorE = end
            }
        }

        ans.add(intArrayOf(cursorS, cursorE))

        return ans.toTypedArray()

    }
}