有很多不错的答案，但我没有找到一个像样的Java实现。以下是我的观点:

public void findNext(int[] nums) {
    int i = nums.length - 1;
    // nums[i - 1] will be the first non increasing number
    while (i > 0 && nums[i] <= nums[i - 1]) {
        i--;
    }
    if (i == 0) {
        System.out.println("it has been the greatest already");
    } else {
        // Find the smallest digit in the second sequence that is larger than it:
        int j = nums.length - 1;
        while (j >= 0 && nums[j] < nums[i - 1]) {
            j--;
        }
        swap(nums, i - 1, j);
        Arrays.sort(nums, i, nums.length);
        System.out.println(Arrays.toString(nums));
    }
}

public void swap(int[] nums, int i, int j) {
    int tmp = nums[i];
    nums[i] = nums[j];
    nums[j] = tmp;
}

下面是生成一个数字的所有排列的代码..不过必须先使用string . valueof (integer)将该整数转换为字符串。

/**
 * 
 * Inserts a integer at any index around string.
 * 
 * @param number
 * @param position
 * @param item
 * @return
 */
public String insertToNumberStringAtPosition(String number, int position,
        int item) {
    String temp = null;
    if (position >= number.length()) {
        temp = number + item;
    } else {
        temp = number.substring(0, position) + item
                + number.substring(position, number.length());
    }
    return temp;
}

/**
 * To generate permutations of a number.
 * 
 * @param number
 * @return
 */
public List<String> permuteNumber(String number) {
    List<String> permutations = new ArrayList<String>();
    if (number.length() == 1) {
        permutations.add(number);
        return permutations;
    }
    // else
    int inserterDig = (int) (number.charAt(0) - '0');
    Iterator<String> iterator = permuteNumber(number.substring(1))
            .iterator();
    while (iterator.hasNext()) {
        String subPerm = iterator.next();
        for (int dig = 0; dig <= subPerm.length(); dig++) {
            permutations.add(insertToNumberStringAtPosition(subPerm, dig,
                    inserterDig));
        }
    }
    return permutations;
}

关于如何做到这一点，请参阅Knuth的“计算机编程艺术:生成所有排列”(.ps.gz)中的“算法L”。

如果你用c++编程，你可以使用next_permutation:

#include <algorithm>
#include <string>
#include <iostream>

int main(int argc, char **argv) {
  using namespace std; 
   string x;
   while (cin >> x) {
    cout << x << " -> ";
    next_permutation(x.begin(),x.end());
    cout << x << "\n";
  }
  return 0;
}

我很确定你的面试官是想委婉地让你说出这样的话:

local number = 564321;

function split(str)
    local t = {};
    for i = 1, string.len(str) do
        table.insert(t, str.sub(str,i,i));
    end
    return t;
end

local res = number;
local i = 1;
while number >= res do
    local t = split(tostring(res));
    if i == 1 then
        i = #t;
    end
    t[i], t[i-1] = t[i-1], t[i];
    i = i - 1;
    res = tonumber(table.concat(t));
end

print(res);

不一定是最有效或最优雅的解决方案，但它在两个循环中解决了所提供的示例，并像他建议的那样一次交换一个数字。

 private static int GetNextHigherNumber(int num)
        {
            //given 38276 return 38627

            string numberstring = num.ToString();

            char[] sNum = numberstring.ToCharArray();

            for (int i = sNum.Length - 1; i > 0; i--)
            {
                for (int j = i - 1; j > 0; j--)
                {
                    if (sNum[i] > sNum[j])
                    {
                        for (int x = i; x > j; x--)
                        {
                            char chr = sNum[x]; 
                            sNum[x] = sNum[x - 1];
                            sNum[x - 1] = chr;
                        }

                        i = 0;
                        break;
                    }
                }
            }

            numberstring = string.Empty;
            for(int x= 0 ; x<sNum.Length;x++)
            {
                numberstring += sNum[x].ToString();
            }

            return Convert.ToInt32(numberstring);
        }