Given a positive 32-bit integer n, you need to find the smallest 32-bit integer which has exactly the same digits existing in the integer n and is greater in value than n. If no such positive 32-bit integer exists, you need to return -1.

Example 1:
Input: 12
Output: 21

Example 2:
Input: 21
Output: -1

Solution：

思路：
Ex. 12443322 ->13222344
0.如果是个纯降序排列的数字，做任何改变都不会使数字变大，直接返回-1
1.从后往前，找找/\到的转折点，也就是Ex中的2，如果从后往前看的话，2是第一个小于其右边位数的数字。再来看如何确定2和谁交换，从后往前遍历，找到第一个大于2的数字交换。
2.然后把转折点之后的数字按升序排列就是最终的结果了。

Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

public class Solution {
    public int nextGreaterElement(int n) {
        char[] number = (n + "").toCharArray();
        
        int i, j;
        // I) Start from the right most digit and 
        // find the first digit that is
        // smaller than the digit next to it.
        for (i = number.length - 1; i > 0; i--) {
            if (number[i - 1] < number[i]) {
               break;
            }
        }

        // If no such digit is found, its the edge case 1.
        if (i == 0)
            return -1;
            
         // II) Find the smallest digit on right side of (i-1)'th 
         // digit that is greater than number[i-1]
        int x = number[i-1], smallest = i;
        for (j = i+1; j < number.length; j++) {
            if (number[j] > x && number[j] <= number[smallest]) {
                smallest = j;
            }
        }
        
        // III) Swap the above found smallest digit with 
        // number[i-1]
        char temp = number[i-1];
        number[i-1] = number[smallest];
        number[smallest] = temp;
        
        // IV) Sort the digits after (i-1) in ascending order
        Arrays.sort(number, i, number.length);
        
        long val = Long.parseLong(new String(number));
        return (val <= Integer.MAX_VALUE) ? (int) val : -1;
    }
}