#include <iostream>
using namespace std;

int main ()
{
  int num=15432;
  int quot,rem;
  int numarr[5];
  int length=0;
  while(num!=0)
  {
      rem=num%10;
      num = num/10;
      numarr[length]=rem;
      length++;
  }

 for(int j=0;j<length;j++)
  {
  for(int i=0;i<length;i++)
  {
      if(numarr[i]<numarr[i+1])
      {
          int tmp=numarr[i];
          numarr[i]=numarr[i+1];
          numarr[i+1]=tmp;
      }
  }
  }

  for(int j=0;j<length;j++)
  {
   cout<<numarr[j];
  }
  return 0;
}

下面是Python中的一个紧凑(但部分是蛮力)解决方案

def findnext(ii): return min(v for v in (int("".join(x)) for x in
    itertools.permutations(str(ii))) if v>ii)

在c++中，你可以这样排列:https://stackoverflow.com/a/9243091/1149664(它与itertools中的算法相同)

以下是Weeble和BlueRaja描述的顶部答案的实现(其他答案)。我怀疑还有什么更好的办法。

def findnext(ii):
    iis=list(map(int,str(ii)))
    for i in reversed(range(len(iis))):
        if i == 0: return ii
        if iis[i] > iis[i-1] :
            break        
    left,right=iis[:i],iis[i:]
    for k in reversed(range(len(right))):
        if right[k]>left[-1]:
           right[k],left[-1]=left[-1],right[k]
           break
    return int("".join(map(str,(left+sorted(right)))))

#include<bits/stdc++.h>
using namespace std;
int main() 
{
    int i,j,k,min,len,diff,z,u=0,f=0,flag=0;
    char temp[100],a[100]`enter code here`,n;
    min=9999;
    //cout<<"Enter the number\n";
    cin>>a;
    len=strlen(a);
    for(i=0;i<len;i++)
    {
        if(a[i]<a[i+1]){flag=1;break;}
    }
    if(flag==0){cout<<a<<endl;}
    else
    {
        for(i=len-1;i>=0;i--)if(((int)a[i-1])<((int)a[i]))break;
        for(k=0;k<i-1;k++)cout<<a[k];
        for(j=i;j<len;j++)
        {
            if(((int)a[j]-48)-((int)a[i-1]-48)>0)
            {
                diff=((int)a[j]-48)-((int)a[i-1]-48);
                if(diff<min){n=a[j];min=diff;}
            }
        }
        cout<<n;
        for(z=i-1;z<len;z++)
        {
            temp[u]=a[z];
            u++;
        }
        temp[u]='\0';
        sort(temp,temp+strlen(temp));
        for(z=0;z<strlen(temp);z++){if(temp[z]==n&&f==0){f=1;continue;}cout<<temp[z];}
    }
    return 0;
}

这是我在Ruby中的实现:

def foo num  
  num = num.to_s.chars.map(&:to_i)
  return num.join.to_i if num.size < 2
  for left in (num.size-2).downto(0) do
    for right in (num.size-1).downto(left+1) do
      if num[right]>num[left]
        num[left],num[right] = num[right],num[left]        
        return (num[0..left] + num[left+1..num.size-1].sort).join.to_i
      end
    end
  end
  return num.join.to_i
end

p foo 38276 
#will print: 38627

import java.util.Scanner;
public class Big {

    public static void main(String[] args) {


        Scanner sc = new Scanner(System.in);
        System.out.print("Enter the number ");
        String str = sc.next();
        int t=0;

        char[] chars  = str.toCharArray();



        for(int i=str.length()-1,j=str.length()-2;j>=0;j--)
        {


                if((int)chars[i]>(int)chars[j])
                {
                    t = (int)chars[i];
                    chars[i] = chars[j];
                    chars[j]=(char)t;

                    for(int k=j+1;k<str.length()-1;k++)
                    {
                        for(int l=k+1;l<str.length();l++)
                        {
                            if(chars[k]>chars[l])
                            {
                                int m = (int)chars[k];
                                chars[k] = chars[l];
                                chars[l]=(char)m;
                            }
                        }
                    }

                    break;
                }






        }
        System.out.print("The next Big number is: ");

        for(int i=0;i<str.length();i++){
            System.out.print(chars[i]);
        }
        sc.close();
    }


}

这里是Java实现

public static int nextHigherNumber(int number) {
    Integer[] array = convertToArray(number);
    int pivotIndex = pivotMaxIndex(array);
    int digitInFirstSequence = pivotIndex -1;
    int lowerDigitIndexInSecondSequence = lowerDigitIndex(array[digitInFirstSequence], array, pivotIndex);
    swap(array, digitInFirstSequence, lowerDigitIndexInSecondSequence);
    doRercursiveQuickSort(array, pivotIndex, array.length - 1);
    return arrayToInteger(array);
}

public static Integer[] convertToArray(int number) {
    int i = 0;
    int length = (int) Math.log10(number);
    int divisor = (int) Math.pow(10, length);
    Integer temp[] = new Integer[length + 1];

    while (number != 0) {
        temp[i] = number / divisor;
        if (i < length) {
            ++i;
        }
        number = number % divisor;
        if (i != 0) {
            divisor = divisor / 10;
        }
    }
    return temp;
}

private static int pivotMaxIndex(Integer[] array) {
    int index = array.length - 1;
    while(index > 0) {
        if (array[index-1] < array[index]) {
            break;
        }
        index--;
    }       
    return index;
}

private static int lowerDigitIndex(int number, Integer[] array, int fromIndex) {
    int lowerMaxIndex = fromIndex;
    int lowerMax = array[lowerMaxIndex];
    while (fromIndex < array.length - 1) {
        if (array[fromIndex]> number && lowerMax > array[fromIndex]) {
            lowerMaxIndex = fromIndex; 
        }
        fromIndex ++;
    }
    return lowerMaxIndex;
}

public static int arrayToInteger(Integer[] array) {
    int number = 0;
    for (int i = 0; i < array.length; i++) {
        number+=array[i] * Math.pow(10, array.length-1-i);
    }
    return number;
}

这里是单元测试

@Test
public void nextHigherNumberTest() {
    assertThat(ArrayUtils.nextHigherNumber(34722641), is(34724126));
    assertThat(ArrayUtils.nextHigherNumber(123), is(132));
}