Leetcode-556. Next Greater Element III

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前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

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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: 12Output: 21

Example 2:

Input: 21Output: -1

这个题目其实可以理解成全排列的下一个。小心一个坑是，下一个排列必须在正整数范围内。

public class Solution {    public int nextGreaterElement(int n) {        String num = n + "";        char[] charArray = num.toCharArray();        if(num.length() < 2) return -1;        int index = num.length() - 2;        for(; index >= 0; index --){            if(num.charAt(index) < num.charAt(index + 1)) break;        }        if(index == -1) return -1;        int swapIndex = index + 1;        for(;swapIndex < num.length();swapIndex ++){            if(num.charAt(swapIndex) <= num.charAt(index)) break;        }        swapIndex --;        char temp = charArray[index];        charArray[index] = charArray[swapIndex];        charArray[swapIndex] = temp;        Arrays.sort(charArray,index + 1, charArray.length);        int result = 0;        for(char ch : charArray) {result = result * 10 + (ch - '0');if(result > Integer.MAX_VALUE / 10) return -1;}                return result;    }}

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