LeetCode in Kotlin
503. Next Greater Element II
Medium
Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums.
The next greater number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn’t exist, return -1 for this number.
Example 1:
Input: nums = [1,2,1]
Output: [2,-1,2] Explanation: The first 1’s next greater number is 2; The number 2 can’t find next greater number. The second 1’s next greater number needs to search circularly, which is also 2.
Example 2:
Input: nums = [1,2,3,4,3]
Output: [2,3,4,-1,4]
Constraints:
1 <= nums.length <= 104-109 <= nums[i] <= 109
Solution
import java.util.ArrayDeque
import java.util.Deque
class Solution {
fun nextGreaterElements(nums: IntArray): IntArray {
val result = IntArray(nums.size)
val stack: Deque<Int> = ArrayDeque()
for (i in nums.size * 2 - 1 downTo 0) {
while (stack.isNotEmpty() && nums[stack.peek()] <= nums[i % nums.size]) {
stack.pop()
}
result[i % nums.size] = if (stack.isEmpty()) -1 else nums[stack.peek()]
stack.push(i % nums.size)
}
return result
}
}