[LeetCode] Problem 503 - Next Greater Element II

Posted on | In ,

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

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.

Code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
public int[] nextGreaterElements(int[] nums) {
    int[] result = new int[nums.length];
    Stack<Integer> stack = new Stack<>();

    Arrays.fill(result, -1);

    for (int i = 0; i < 2 * nums.length; i++) {
        int idx = i % nums.length;

        while (!stack.isEmpty() && nums[stack.peek()] < nums[idx])
            result[stack.pop()] = nums[idx];

        if (i < nums.length)
            stack.push(i);
    }

    return result;
}