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.

解题要点：

声明一个长度为两倍原始数组的array（作为平铺版的circular数组），用迭代的方法，在原始数组里找出此后第一个大于它的数。

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int[] res = new int[nums.length];
        int[] doublenums = new int[2 * nums.length];
        for(int k = 0; k < nums.length; k++){
            doublenums[k] = nums[k];
        }
        for(int p = nums.length; p < doublenums.length; p++){
            doublenums[p] = nums[p - nums.length];
        }
        for(int i = 0; i < nums.length; i++){
            res[i] = -1;
            for(int j = i + 1; j < doublenums.length; j++){
                if(doublenums[i] < doublenums[j]){
                    res[i] = doublenums[j];
                    break;
                }
            }
        }
        return res;
    }
}