503 Next Greater Element II

503. Next Greater Element II

1. Question

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.

2. Implementation

思路：这道题和Next Greater Element I的区别就在于这里数组是一个环，所以数组最后的数组的Greater Next Element可能在它前面。方法就是拆环成线，另外和一个不同点是Next Greater Element I的输入是两个数组，而这里只有一个数组，所以stack存的是数组的Index而不是数组的元素

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int n = nums.length;
        int[] res = new int[n];
        Arrays.fill(res, -1);

        Stack<Integer> stack = new Stack<>();

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

            if (i < n) {
                stack.push(i);
            }
        }

        return res;
    }
}

3. Time & Space Complexity

时间和空间复杂度都是O(n)