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.

题目含义：给一个循环数组，返回一个等长的数组，数组中的每一个元素是：它后面的第一个大于它的元素（如果后面没有就循环一遍到最前面找，直到循环了一圈为止），如果不存在这样的数，就返回-1～

class Solution {
    private int getNextGreater(int[] nums,int begin,int end,int target)
    {
        if (begin>nums.length -1 || end>nums.length-1 || begin > end) return Integer.MIN_VALUE;
        for (int i=begin;i<=end;i++)
        {
            if (nums[i] > target) return nums[i];
        }
        return Integer.MIN_VALUE;
    }    
    
    public int[] nextGreaterElements(int[] nums) {
        int[] result = new int[nums.length];
        if (nums.length==0) return new int[]{};
        if (nums.length ==1) return new int[]{-1};
        for (int i=0;i<nums.length;i++)
        {
            int greaterLeft = Integer.MIN_VALUE;
            int greaterRight = Integer.MIN_VALUE;
            if (i==0)
            {
                greaterRight = getNextGreater(nums,1,nums.length-1,nums[i]);
            }else if (i == nums.length-1)
            {
                greaterLeft = getNextGreater(nums,0,i-1,nums[i]);
            }else
            {
                greaterRight = getNextGreater(nums,i+1,nums.length-1,nums[i]);
                if (greaterRight==Integer.MIN_VALUE)
                {
                    greaterLeft = getNextGreater(nums,0,i-1,nums[i]);
                }
            }
            if (greaterRight==Integer.MIN_VALUE && greaterLeft==Integer.MIN_VALUE) result[i] =-1;
            else  result[i] = Math.max(greaterLeft,greaterRight);
        }
        return result;    
    }
}