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.

Solution1：

思路：
Time Complexity: O(N^2) Space Complexity: O(N)

Solution2：

思路：
The approach is same as Next Greater Element I
The only difference here is that we use stack to keep the indexes of the decreasing subsequence
Time Complexity: O(N) Space Complexity: O(N)

Solution1 Code：

Solution2 Code：

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int n = nums.length, next[] = new int[n];
        Arrays.fill(next, -1);
        Stack stack = new Stack<>(); // index stack
        for (int i = 0; i < n * 2; i++) {
            int num = nums[i % n]; 
            while (!stack.isEmpty() && nums[stack.peek()] < num)
                next[stack.pop()] = num;
            if (i < n) stack.push(i);
        }   
        return next;
    }
}