Lintcode 1201. 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

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.

Example 2:

Input: [1]
Output: [-1]
Explanation: 
The number 1 can't find next greater number.

Notice

The length of given array won't exceed 10000.

Solution:

monotonic descreasing stack. 

Code (Java):

public class Solution {
    /**
     * @param nums: an array
     * @return: the Next Greater Number for every element
     */
    public int[] nextGreaterElements(int[] nums) {
        // Write your code here
        if (nums == null || nums.length == 0) {
            return new int[0];
        }
        
        int[] ans = new int[nums.length];
        
        // init to -1
        for (int i = 0; i < nums.length; i++) {
            ans[i] = -1;
        }
        
        // stack is mono descreasing stack, and it stores the index
        Stack<Integer> stack = new Stack<>();
        
        for (int i = 0; i < nums.length * 2; i++) {
            int ii = i % nums.length;
            while (!stack.isEmpty() && nums[ii] > nums[stack.peek()]) {
                ans[stack.pop()] = nums[ii];
            }
            
            stack.push(ii);
        }
        
        return ans;
        
    }
}

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int[] ans = new int[nums.length];
        
        Stack<Integer> stack = new Stack<>();
        
        for (int i = nums.length * 2 - 1; i>= 0; i--) {
            int ii = i % nums.length;
            
            while (!stack.isEmpty() && nums[ii] >= nums[stack.peek()]) {
                stack.pop();
            }
            
            ans[ii] = stack.isEmpty() ? -1 : nums[stack.peek()];
            
            stack.push(ii);
        }
        
        return ans;
    }
}