原题链接在这里：https://leetcode.com/problems/next-greater-element-iii/description/
题目：
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
题解：
如果每一位digit依次递减, 比如"54321"就没有next greater element. return -1.
所以说节点在于不是递减的位置, "52431", 找到 “2” 和 “4”的位置不是递减. 然后在2的后面比2大的最小digit, 这里是3.
"2", "3"换位, 变成"53421", 再把3后面的部分sort成从小到大 "53124"就是next greater element.
Time Complexity: O(x), x是n的digit 数目. 每个digit不会走超过3遍.
Space: O(x).
AC Java:
1 class Solution { 2 public int nextGreaterElement(int n) { 3 char [] arr = (""+n).toCharArray(); 4 int i = arr.length-2; 5 while(i>=0 && arr[i]>=arr[i+1]){ 6 i--; 7 } 8 9 if(i < 0){ 10 return -1; 11 } 12 13 int j = arr.length-1; 14 while(j>i && arr[j]<=arr[i]){ 15 j--; 16 } 17 18 swap(arr, i, j); 19 reverse(arr, i+1); 20 try{ 21 return Integer.valueOf(new String(arr)); 22 }catch(Exception e){ 23 return -1; 24 } 25 } 26 27 private void swap(char [] arr, int i, int j){ 28 char temp = arr[i]; 29 arr[i] = arr[j]; 30 arr[j] = temp; 31 } 32 33 private void reverse(char [] arr, int start){ 34 int i = start; 35 int j = arr.length-1; 36 while(i < j){ 37 swap(arr, i++, j--); 38 } 39 } 40 }
类似Next Permutation, Next Greater Element I, Next Greater Element II.