LeetCode Next Greater Element III

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:

 class Solution {
     public int nextGreaterElement(int n) {
         char [] arr = (""+n).toCharArray();
         int i = arr.length-2;
         while(i>=0 && arr[i]>=arr[i+1]){
             i--;
         }

         if(i < 0){
             return -1;
         }

         int j = arr.length-1;
         while(j>i && arr[j]<=arr[i]){
             j--;
         }

         swap(arr, i, j);
         reverse(arr, i+1);
         try{
             return Integer.valueOf(new String(arr));
         }catch(Exception e){
             return -1;
         }
     }

     private void swap(char [] arr, int i, int j){
         char temp = arr[i];
         arr[i] = arr[j];
         arr[j] = temp;
     }

     private void reverse(char [] arr, int start){
         int i = start;
         int j = arr.length-1;
         while(i < j){
             swap(arr, i++, j--);
         }
     }
 }

类似Next Permutation, Next Greater Element I, Next Greater Element II.