66. Next Greater Element II

Topic :

stack

Difficulty :

medium

Problem Link :

problem statement

Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums.

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, return -1 for this number.

Example 1:

Input: nums = [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: nums = [1,2,3,4,3]
Output: [2,3,4,-1,4]

Constraints:

1 <= nums.length <= 104
-109 <= nums[i] <= 109

solution

import java.io.*;
import java.util.*;
 class NextGreaterElement_II
{
  public static void main(String args[]){
      int nums[]={1,2,1};
      System.out.println(Arrays.toString(nextGreaterElements(nums)));
  }
  static int[] nextGreaterElements(int[] nums) {
        int ans[]=new int[nums.length];
        Stack<Integer> stack= new Stack<>();
        for(int i=nums.length-1;i>=0;i--)
            stack.push(nums[i]);
        
       
        for(int i=nums.length-1;i>=0;i--){
            int ele=nums[i];
            while(!stack.isEmpty() && stack.peek()<=ele)
                stack.pop();
            if(stack.isEmpty())
                ans[i]=-1;
            else
                ans[i]=stack.peek();
            stack.push(nums[i]);
        }
        return ans;
        
    }
}