503. Next Greater Element II - LeetCode

Description:

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.

Accepted

58,855

Submissions

112,982

 

Solution:

 

 

 

 

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        
        boolean flag=false;
        
        int[] res = new int[nums.length];
        
        for(int i = 0 ; i<nums.length; i++){
            flag = false; 
            for(int k = i+1; k<nums.length; k++){
                
                if(nums[k]>nums[i]){
                     //System.out.println(" greater "+ nums[k]);
                    res[i] = nums[k];
                    flag = true;
                    break;
                }
                
                
            }
            
            if(!flag){
                for(int m = 0; m<i; m++){
                    
                    if(nums[m]>nums[i])
                    {
                       //System.out.println(" greater 2 "+ nums[m] );
                        res[i] = nums[m];
                        flag = true;
                        break;
                    }
                
                } 
            }
           
           if(!flag){
                //System.out.println(" -1 ");
               res[i] = -1;
           }
        }
        
        return res;
    }
}