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第二道题答案（程序），供参考。

mike20030405(麦客)
本文发表在 rolia.net 枫下论坛#include <stdio.h>

int print_array(int *input, int number)
{
int i;

printf("number = %d [", number);
for(i=0; i<number; i++)
{
printf("%d ", *input);
input++;
}
printf("]\n");
}

//return:
// -1 - empty array;
// 1 - no even inetegers found in the array;
// even - the multiplied value;

int seek_array(int *input, int number)
{
int i;
int largest_even = 9999;
int the_less_even = 9997;

if (number <= 0 )
return -1;

for(i=0; i<number; i++, input++)
{
if (((*input)/2)*2 != *input) continue;

if (largest_even == 9999 && the_less_even == 9997)
largest_even = *input;
else if (the_less_even == 9997)
{
if (*input <= largest_even)
the_less_even = *input;
else
{
the_less_even = largest_even;
largest_even = *input;
}
}
else
{
if (*input < the_less_even)
;
else if (*input == the_less_even)
;
else if (*input > largest_even)
{
the_less_even = largest_even;
largest_even = *input;
}
else if (*input == largest_even)
the_less_even = largest_even;
else //in case that *input less than the_less_even;
the_less_even = *input;
}
}

printf("largest_even = %d, the_less_even=%d\n", largest_even, the_less_even);


if (largest_even == 9999 && the_less_even == 9997)
return 1;
else if (the_less_even == 9997)
return largest_even;
else
return largest_even*the_less_even;
}

int main()
{
// int int_array[]={};
// int int_array[]={3};
// int int_array[]={4};
// int int_array[]={0, 0};
// int int_array[]={12, 10};
// int int_array[]={0, 0, 2};
// int int_array[]={135, 51, 92147};
// int int_array[]={1, 2, 3, 5, -4, 9, -8};
int int_array[]={0, 1, 0, 2, 3, 4, 5, 10, 5, 20, -2, 9};
int return_value=0;

print_array(&int_array[0], sizeof(int_array)/sizeof(int));
return_value = seek_array(&int_array[0], sizeof(int_array)/sizeof(int));
printf("seek_array() returned: %d\n", return_value);
}更多精彩文章及讨论，请光临枫下论坛 rolia.net
(#3600441@0)
Last Updated: 2007-4-7
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Replies, comments and Discussions:

  • 工作学习 / 专业技术讨论 / 微软面试题求解， 多谢。 -tonyhao(tonyhao); 2007-4-5 {598} (#3596764@0)
    1. Suppose you have an integer array a of size n that is ordered but has duplicates. For example a = [ 0, 0, 0, 2, 3, 3, 5, 5, 5, 7, 9, 9] (n = 12). Write a function that removes the duplicates, and then returns the new size. In above example, a becomes [ 0, 2, 3, 5, 7, 9 ] and the function returns 6. Do it in-place, meaning no extra memory space should be allocated to store intermediate or final results.

    2. Write a function that will accept an array of integers and the length of the array, find the two largest even integers in the array, multiply them together, and return the result.
    • C语言的大侠都跑哪去了？ -tonyhao(tonyhao); 2007-4-5 (#3596814@0)
    • Your question 1 remove((char*)myints+*(unique( myints, myints+sizeof(myints)/sizeof(int) ))+2); -interview(intervieweree); 2007-4-5 (#3596990@0)
      • remove((char*)myints+*(unique( myints, myints+sizeof(myints)/sizeof(int) ))+2); -interview(intervieweree); 2007-4-5 (#3596992@0)
      • 没看懂， 能写成程序吗？ -tonyhao(tonyhao); 2007-4-5 (#3597025@0)
        • by using STL functions, you dont see a interim value in your code. I think this is the purpose of the test. Do you really believe no interim value will be used to swap values? -interview(intervieweree); 2007-4-5 (#3597039@0)
          • sorry, swap value does not need intermediate variable. -interview(intervieweree); 2007-4-5 (#3597046@0)
          • 微软都是让用C语言写， 不能用STL -tonyhao(tonyhao); 2007-4-5 (#3597050@0)
    • 要求用标准C语言写。 -tonyhao(tonyhao); 2007-4-5 (#3597116@0)
    • Q1: -shz(shz); 2007-4-5 {224} (#3597967@0)
      int UniqueArray(int *arr, int size)
      {
      int i, j = 1;

      for (i = 1; i < size; i++)
      {
      if (arr[j-1] != arr[i])
      {
      arr[j] = arr[i];
      j++;
      }
      }
      return j;
      }
      • what if size==0? and if there is no duplication, all elements are unnecessarily assigned, so arr[j] = arr[i]; is needed only when i!=j -wangqingshui(忘情水); 2007-4-5 (#3598034@0)
    • 第一道题的答案（程序）见内，供参考。 -mike20030405(麦客); 2007-4-7 {1586} (#3600320@0)
      本文发表在 rolia.net 枫下论坛#include <stdio.h>

      int print_array(int *input, int number)
      {
      int i;

      printf("number = %d [", number);
      for(i=0; i<number; i++)
      {
      printf("%d ", *input);
      input++;
      }
      printf("]\n");
      }

      int re_order(int *input, int number)
      {
      int i;
      int final_number = 0;
      int available_position = 0;
      int offset = 0;

      if (number <= 1)
      return number;

      final_number = 1;
      available_position++;
      offset++;

      for(i=1; i<number; i++)
      {
      if (*(input+offset) == *(input+available_position-1))
      {
      *(input+offset) = 0;
      offset++;
      }
      else
      {
      final_number++;
      *(input+available_position) = *(input+offset);
      available_position++;
      offset++;
      }

      }

      return final_number;
      }

      int main()
      {
      // int int_array[]={};
      // int int_array[]={0};
      // int int_array[]={0, 0};
      // int int_array[]={0, 0, 2};
      int int_array[]={0, 0, 0, 2, 3, 3, 5, 5, 5, 7, 9, 9};
      int new_number;

      print_array(&int_array[0], sizeof(int_array)/sizeof(int));
      new_number = re_order(&int_array[0], sizeof(int_array)/sizeof(int));
      print_array(&int_array[0], new_number);
      }更多精彩文章及讨论，请光临枫下论坛 rolia.net
      • 第13行如果改为printf("]\n ");输出会更清楚些。 -mike20030405(麦客); 2007-4-7 (#3600327@0)
      • I prefer input[offset] over *(input+offset) for better readability. If I'd like to increase performance, I'd replace *(input+offset) with *next (of type int*) and ++offset with ++next. -wangqingshui(忘情水); 2007-4-8 (#3600954@0)
        • good point! -mike20030405(麦客); 2007-4-8 (#3601888@0)
    • 第二道题答案（程序），供参考。 -mike20030405(麦客); 2007-4-7 {2993} (#3600441@0)
      本文发表在 rolia.net 枫下论坛#include <stdio.h>

      int print_array(int *input, int number)
      {
      int i;

      printf("number = %d [", number);
      for(i=0; i<number; i++)
      {
      printf("%d ", *input);
      input++;
      }
      printf("]\n");
      }

      //return:
      // -1 - empty array;
      // 1 - no even inetegers found in the array;
      // even - the multiplied value;

      int seek_array(int *input, int number)
      {
      int i;
      int largest_even = 9999;
      int the_less_even = 9997;

      if (number <= 0 )
      return -1;

      for(i=0; i<number; i++, input++)
      {
      if (((*input)/2)*2 != *input) continue;

      if (largest_even == 9999 && the_less_even == 9997)
      largest_even = *input;
      else if (the_less_even == 9997)
      {
      if (*input <= largest_even)
      the_less_even = *input;
      else
      {
      the_less_even = largest_even;
      largest_even = *input;
      }
      }
      else
      {
      if (*input < the_less_even)
      ;
      else if (*input == the_less_even)
      ;
      else if (*input > largest_even)
      {
      the_less_even = largest_even;
      largest_even = *input;
      }
      else if (*input == largest_even)
      the_less_even = largest_even;
      else //in case that *input less than the_less_even;
      the_less_even = *input;
      }
      }

      printf("largest_even = %d, the_less_even=%d\n", largest_even, the_less_even);


      if (largest_even == 9999 && the_less_even == 9997)
      return 1;
      else if (the_less_even == 9997)
      return largest_even;
      else
      return largest_even*the_less_even;
      }

      int main()
      {
      // int int_array[]={};
      // int int_array[]={3};
      // int int_array[]={4};
      // int int_array[]={0, 0};
      // int int_array[]={12, 10};
      // int int_array[]={0, 0, 2};
      // int int_array[]={135, 51, 92147};
      // int int_array[]={1, 2, 3, 5, -4, 9, -8};
      int int_array[]={0, 1, 0, 2, 3, 4, 5, 10, 5, 20, -2, 9};
      int return_value=0;

      print_array(&int_array[0], sizeof(int_array)/sizeof(int));
      return_value = seek_array(&int_array[0], sizeof(int_array)/sizeof(int));
      printf("seek_array() returned: %d\n", return_value);
      }更多精彩文章及讨论，请光临枫下论坛 rolia.net
      • 2个int的最大偶数，相乘后不是int了吧， 该用long吧 -tonyhao(tonyhao); 2007-4-8 (#3601106@0)
        • good point!! -mike20030405(麦客); 2007-4-8 (#3601889@0)
      • 2 minor concerns: 1. operator % can be used. 2. sequence of conditions can be reordered to ommit no action conditions like if (*input < the_less_even) ; -wangqingshui(忘情水); 2007-4-8 (#3601127@0)
        • good point! -mike20030405(麦客); 2007-4-8 (#3601890@0)

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