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。
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
这是个很有趣的问题。
这是我的java版本。在我检查其他贡献者的评论之前,从弄清楚模式到完全完成代码,我花了大约3个小时。很高兴看到我的想法和别人一样。
O (n)的解决方案。老实说,如果时间只有15分钟,并且要求在白板上完成完整的代码,我将会失败。
以下是我的解决方案的一些有趣点:
避免任何排序。 完全避免字符串操作 实现O(logN)空间复杂度
我在代码中添加了详细注释,并在每个步骤中添加了大O。
public int findNextBiggestNumber(int input ) {
//take 1358642 as input for example.
//Step 1: split the whole number to a list for individual digital 1358642->[2,4,6,8,5,3,1]
// this step is O(n)
int digitalLevel=input;
List<Integer> orgNumbersList=new ArrayList<Integer>() ;
do {
Integer nInt = new Integer(digitalLevel % 10);
orgNumbersList.add(nInt);
digitalLevel=(int) (digitalLevel/10 ) ;
} while( digitalLevel >0) ;
int len= orgNumbersList.size();
int [] orgNumbers=new int[len] ;
for(int i=0;i<len;i++){
orgNumbers[i ] = orgNumbersList.get(i).intValue();
}
//step 2 find the first digital less than the digital right to it
// this step is O(n)
int firstLessPointer=1;
while(firstLessPointer<len&&(orgNumbers[firstLessPointer]>orgNumbers[ firstLessPointer-1 ])){
firstLessPointer++;
}
if(firstLessPointer==len-1&&orgNumbers[len-1]>=orgNumbers[len-2]){
//all number is in sorted order like 4321, no answer for it, return original
return input;
}
//when step 2 step finished, firstLessPointer pointing to number 5
//step 3 fristLessPointer found, need to find to first number less than it from low digital in the number
//This step is O(n)
int justBiggerPointer= 0 ;
while(justBiggerPointer<firstLessPointer&& orgNumbers[justBiggerPointer]<orgNumbers[firstLessPointer]){
justBiggerPointer++;
}
//when step 3 finished, justBiggerPointer pointing to 6
//step 4 swap the elements of justBiggerPointer and firstLessPointer .
// This is O(1) operation for swap
int tmp= orgNumbers[firstLessPointer] ;
orgNumbers[firstLessPointer]= orgNumbers[justBiggerPointer] ;
orgNumbers[justBiggerPointer]=tmp ;
// when step 4 finished, the list looks like [2,4,5,8,6,3,1] the digital in the list before
// firstLessPointer is already sorted in our previous operation
// we can return result from this list but in a differrent way
int result=0;
int i=0;
int lowPointer=firstLessPointer;
//the following pick number from list from the position just before firstLessPointer, here is 8 -> 5 -> 4 -> 2
//This Operation is O(n)
while(lowPointer>0) {
result+= orgNumbers[--lowPointer]* Math.pow(10,i);
i++;
}
//the following pick number from list from position firstLessPointer
//This Operation is O(n)
while(firstLessPointer<len) {
result+= orgNumbers[firstLessPointer++ ]* Math.pow(10,i);
i++;
}
return result;
}
下面是在Intellj中运行的结果:
959879532-->959892357
1358642-->1362458
1234567-->1234576
77654321-->77654321
38276-->38627
47-->74
其他回答
#include<stdio.h>
#include<cstring>
#include<iostream>
#include<string.h>
#include<sstream>
#include<iostream>
using namespace std;
int compare (const void * a, const void * b)
{
return *(char*)a-*(char*)b;
}
/*-----------------------------------------------*/
int main()
{
char number[200],temp;
cout<<"please enter your number?"<<endl;
gets(number);
int n=strlen(number),length;
length=n;
while(--n>0)
{
if(number[n-1]<number[n])
{
for(int i=length-1;i>=n;i--)
{
if(number[i]>number[n-1])
{
temp=number[i];
number[i]=number[n-1];
number[n-1]=temp;
break;
}
}
qsort(number+n,length-n,sizeof(char),compare);
puts(number);
return 0;
}
}
cout<<"sorry itz the greatest one :)"<<endl;
}
只是使用python的另一个解决方案:
def PermutationStep(num):
if sorted(list(str(num)), reverse=True) == list(str(num)):
return -1
ls = list(str(num))
n = 0
inx = 0
for ind, i in enumerate(ls[::-1]):
if i < n:
n = i
inx = -(ind + 1)
break
n = i
ls[inx], ls[inx + 1] = ls[inx + 1], ls[inx]
nl = ls[inx::-1][::-1]
ln = sorted(ls[inx+1:])
return ''.join(nl) + ''.join(ln)
print PermutationStep(23514)
输出:
23541
这是我的代码,它是这个例子的修改版本
库:
class NumPermExample
{
// print N! permutation of the characters of the string s (in order)
public static void perm1(String s, ArrayList<String> perm)
{
perm1("", s);
}
private static void perm1(String prefix, String s, ArrayList<String> perm)
{
int N = s.length();
if (N == 0)
{
System.out.println(prefix);
perm.add(prefix);
}
else
{
for (int i = 0; i < N; i++)
perm1(prefix + s.charAt(i), s.substring(0, i)
+ s.substring(i+1, N));
}
}
// print N! permutation of the elements of array a (not in order)
public static void perm2(String s, ArrayList<String> perm)
{
int N = s.length();
char[] a = new char[N];
for (int i = 0; i < N; i++)
a[i] = s.charAt(i);
perm2(a, N);
}
private static void perm2(char[] a, int n, ArrayList<String> perm)
{
if (n == 1)
{
System.out.println(a);
perm.add(new String(a));
return;
}
for (int i = 0; i < n; i++)
{
swap(a, i, n-1);
perm2(a, n-1);
swap(a, i, n-1);
}
}
// swap the characters at indices i and j
private static void swap(char[] a, int i, int j)
{
char c;
c = a[i]; a[i] = a[j]; a[j] = c;
}
// next higher permutation
public static int nextPermutation (int number)
{
ArrayList<String> perm = new ArrayList<String>();
String cur = ""+number;
int nextPerm = 0;
perm1(cur, perm);
for (String s : perm)
{
if (Integer.parseInt(s) > number
&& (nextPerm == 0 ||
Integer.parseInt(s) < nextPerm))
{
nextPerm = Integer.parseInt(s);
}
}
return nextPerm;
}
}
测试:
public static void main(String[] args)
{
int a = 38276;
int b = NumPermExample.nextPermutation(a);
System.out.println("a: "+a+", b: "+b);
}
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)))))
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; }