Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
645 views
in Technique[技术] by (71.8m points)

algorithm - Good C++ solutions to the "Bring all the zeros to the back of the array" interview challenge

I had an interview for a Jr. development job and he asked me to write a procedure that takes an array of ints and shoves the zeroes to the back. Here are the constraints (which he didn't tell me at the beginning .... As often happens in programming interviews, I learned the constraints of the problem while I solved it lol):

  • Have to do it in-place; no creating temporary arrays, new arrays, etc.
  • Don't have to preserve the order of the nonzero numbers (I wish he would've told me this at the beginning)

Setup:

int arr[] = {0, -2, 4, 0, 19, 69}; 
/* Transform arr to {-2, 4, 19, 69, 0, 0} or {69, 4, -2, 19, 0, 0} 
   or anything that pushes all the nonzeros to the back and keeps
   all the nonzeros in front */

My answer:

bool f (int a, int b) {return a == 0;}
std::sort(arr, arr+sizeof(arr)/sizeof(int), f);

What are some other good answers?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Maybe the interviewer was looking for this answer:

#include <algorithm>
//...
std::partition(std::begin(arr), std::end(arr), [](int n) { return n != 0; });

If the order needs to be preserved, then std::stable_partition should be used:

#include <algorithm>
//...
std::stable_partition(std::begin(arr), std::end(arr), [](int n) { return n != 0; });

For pre C++11:

#include <functional>
#include <algorithm>
//...
std::partition(arr, arr + sizeof(arr)/sizeof(int), 
               std::bind1st(std::not_equal_to<int>(), 0));

Live Example

Basically, if the situation is that you need to move items that satisfy a condition to "one side" of a container, then the partition algorithm functions should be high up on the list of solutions to choose (if not the solution to use).


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...