我刚刚搞砸了一次面试，而且我的面试问题几乎毫无进展。
给定一个数，找出下一个与之完全相同的更大的数 作为原始数字的数字集。例如:给定38276返回 38627
我想从找到第一个比个位小的数字的下标开始。然后我将旋转子集中的最后一个数字，这样它是由相同的数字组成的下一个最大的数字，但卡住了。
面试官还建议我试着一次换一个数字，但我搞不懂算法，只能盯着屏幕看20-30分钟。不用说，我想我还得继续找工作。
当前回答
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);
}
其他回答
回答在java与一个更多的条件添加
Next number should also be an Even number public static int nextDigit(int number) { String num = String.valueOf(number); int stop = 0; char[] orig_chars = null; char[] part1 = null; char[] part2 = null; orig_chars = num.toCharArray(); System.out.println("vivek c r"); for (int i = orig_chars.length - 1; i > 0; i--) { String previous = orig_chars[i - 1] + ""; String next = orig_chars[i] + ""; if (Integer.parseInt(previous) < Integer.parseInt(next)) { if (Integer.parseInt(previous) % 2 == 0) { String partString1 = ""; String partString2 = ""; for (int j = 0; j <= i - 1; j++) { partString1 = partString1.concat(orig_chars[j] + ""); } part1 = partString1.toCharArray(); for (int k = i; k < orig_chars.length; k++) { partString2 = partString2.concat(orig_chars[k] + ""); } part2 = partString2.toCharArray(); Arrays.sort(part2); for (int l = 0; l < part2.length; l++) { char temp = '0'; if (part2[l] > part1[i - 1]) { temp = part1[i - 1]; part1[i - 1] = part2[l]; part2[l] = temp; break; } } for (int m = 0; m < part2.length; m++) { char replace = '0'; if (part2[m] % 2 == 0) { replace = part2[m]; for (int n = m; n < part2.length - 1; n++) { part2[n] = part2[n + 1]; } part2[part2.length - 1] = replace; break; } } System.out.print(part1); System.out.println(part2); System.exit(0); } } } System.out.println("NONE"); return 0; }
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();
}
}
PHP代码
function NextHigherNumber($num1){
$num = strval($num1);
$max = 0;
for($i=(strlen($num)-2); $i>=0; $i--){
$numArrayRaw = substr($num, $i);
$numArray = str_split($numArrayRaw);
$max = max($numArray);
if ($numArray[0] < $max){
sort( $numArray, SORT_NUMERIC );
array_pop($numArray);
$numarrstr = implode("",$numArray);
$rt = substr($num,0,$i) . $max . $numarrstr;
return $rt;
}
}
return "-1";
}
echo NextHigherNumber(123);
我知道这是一个非常老的问题，但我仍然没有在c#中找到简单的代码。这可能会对参加面试的男士有所帮助。
class Program
{
static void Main(string[] args)
{
int inputNumber = 629;
int i, currentIndexOfNewArray = 0;
int[] arrayOfInput = GetIntArray(inputNumber);
var numList = arrayOfInput.ToList();
int[] newArray = new int[arrayOfInput.Length];
do
{
int temp = 0;
int digitFoundAt = 0;
for (i = numList.Count; i > 0; i--)
{
if (numList[i - 1] > temp)
{
temp = numList[i - 1];
digitFoundAt = i - 1;
}
}
newArray[currentIndexOfNewArray] = temp;
currentIndexOfNewArray++;
numList.RemoveAt(digitFoundAt);
} while (arrayOfInput.Length > currentIndexOfNewArray);
Console.WriteLine(GetWholeNumber(newArray));
Console.ReadKey();
}
public static int[] GetIntArray(int num)
{
IList<int> listOfInts = new List<int>();
while (num > 0)
{
listOfInts.Add(num % 10);
num = num / 10;
}
listOfInts.Reverse();
return listOfInts.ToArray();
}
public static double GetWholeNumber(int[] arrayNumber)
{
double result = 0;
double multiplier = 0;
var length = arrayNumber.Count() - 1;
for(int i = 0; i < arrayNumber.Count(); i++)
{
multiplier = Math.Pow(10.0, Convert.ToDouble(length));
result += (arrayNumber[i] * multiplier);
length = length - 1;
}
return result;
}
}
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。
