Next Greater Element II

LeetCode Q 503 - Next Greater Element II

Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. 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, output -1 for this number.

Example 1: Input: [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.
Note: The length of given array won’t exceed 10000.

Solution

Code

public int[] nextGreaterElements(int[] nums) {
  if (nums == null || nums.length == 0)
    return nums;

  Stack<Integer> stack = new Stack<>(); // store indexes

  int len = nums.length;
  int[] res = new int[len];
  Arrays.fill(res, -1);

  for (int i = 0; i < 2 * len; i++) {
    while (!stack.isEmpty() && nums[stack.peek()] < nums[i % len]) {
      res[stack.pop()] = nums[i % len];
    }
    stack.push(i % len);
  }

  return res;
}