556 Next Greater Element III

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 nand 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

题目含义：给定一个正32位整数n，您需要找到大于n但最接近与n的32位整数，它与整数n中存在完全相同的数字。如果没有这样的正32位整数存在，则需要返回-1。

 1     public int nextGreaterElement(int n) {
 2         char[] digs = (n + "").toCharArray();
 3         int one = digs.length - 2;//从后往前找到第一个小于后面一位数字的数字
 4         while (one >= 0) {
 5             if (digs[one] < digs[one + 1]) break;
 6             one--;
 7         }
 8 
 9         if (one < 0) return -1;//没有找到，说明当前数字就是最大的，直接返回-1
10         char minNuber = digs[one];
11         int smallest = one + 1;//找到one后面，比one值大的集合中最小的一位。互换这一位和one位，可以得到最终结果
12         for (int j = one + 1; j < digs.length; j++) {
13             if (digs[j] > minNuber && digs[j] <= digs[smallest])
14                 smallest = j;
15         }
16 
17         char temp = digs[smallest];
18         digs[smallest] = digs[one];
19         digs[one] = temp;
20 
21         Arrays.sort(digs, one + 1, digs.length); //把one后面的所有数位由低到高排列
22         long value = Long.parseLong(new String(digs));
23         return value <= Integer.MAX_VALUE ? (int) value : -1;    
24     }