我如何显示所有 3 个值，如一个低索引、一个高索引和最后一个是最高的，但找不到？

Question

在迭代过程中实现二分查找。一个一个地搜索 3 个值，第一个将放置在低索引和中索引中，第二个将放置在中索引和高索引中，最后一个值将是数组的最大值并且不会找到了。

#include <stdio.h>
int main()
{
  int c, first, last, middle, n, search, array[100];

  printf("Enter number of elements\n");
  scanf("%d", &n);

  printf("Enter %d integers\n", n);

  for (c = 0; c < n; c++)
    scanf("%d", &array[c]);

  printf("Enter first value to find\n");
  scanf("%d", &search);

  printf("Enter second value to find\n");
  scanf("%d", &search);

  printf("Enter third value to find\n");
  scanf("%d", &search);


  first = 0;
  last = n - 1;
  middle = (first+last)/2;

  while (first <= last) {
    if (array[middle] < search)
      first = middle + 1;
    else if (array[middle] == search) {
      printf("%d found at location %d.\n", search, middle+1);
      break;
    }
    else
      last = middle - 1;

    middle = (first + last)/2;
  }
  if (first > last)
    printf("Not found! %d isn't present in the list.\n", search);

  return 0;
}

Answer 1

正确的方法是将应针对不同值重复的代码放在循环或函数中。在这里你可以很容易地使用一个循环：

#include <stdio.h>
int main()
{
    int c, first, last, middle, n, search, array[100];
    const char* label[] = { "first", "second", "third" };

    printf("Enter number of elements\n");
    scanf("%d", &n);

    printf("Enter %d integers\n", n);

    for (c = 0; c < n; c++)
        scanf("%d", &array[c]);

    // idiomatic way for an array length
    unsigned repeat = sizeof(label) / sizeof(*label);
    for (unsigned count = 0; count < repeat; count++) {

        printf("Enter %s value to find\n", label[count]);
        scanf("%d", &search);

        first = 0;
        last = n - 1;
        middle = (first + last) / 2;

        while (first <= last) {
            if (array[middle] < search)
                first = middle + 1;
            else if (array[middle] == search) {
                printf("%d found at location %d.\n", search, middle + 1);
                break;
            }
            else
                last = middle - 1;

            middle = (first + last) / 2;
        }
        if (first > last)
            printf("Not found! %d isn't present in the list.\n", search);
    }

    return 0;
}

但是您应该始终控制输入函数 (scanf) 的 return 值：如果您键入非数字字符（例如 a) 以下所有 scanf 将 return 0 而不是 1 并保持变量不变。然后，您的代码将使用未初始化的值来调用未定义的行为（C 程序员的地狱...）。

你所有的 scanf 调用应该如下所示：

if (1 != scanf("%d", &n)) {
    // process error
    fprintf(stderr, "Incorrect input\n");
    exit(1);
}

您将以有序的方式停止程序并警告用户原因，而不是冒着崩溃或意外结果而没有错误消息的风险。

Answer 2

这里您实际上是在搜索最后一个值。查看代码的以下部分：

  printf("Enter first value to find\n");
  scanf("%d", &search);

  printf("Enter second value to find\n");
  scanf("%d", &search);

  printf("Enter third value to find\n");
  scanf("%d", &search);

您使用同一个变量来存储所有三个整数。因此，在您每次输入之后，搜索的值都会被当前输入所取代。如果您想对所有搜索输入使用单个变量。然后你可以在循环中调用它。请参阅下面的示例以更好地理解：

#include <stdio.h>
int main()
{
  int c, first, last, middle, n, search, array[100];

  printf("Enter number of elements\n");
  scanf("%d", &n);

  printf("Enter %d integers\n", n);

  for (c = 0; c < n; c++)
    scanf("%d", &array[c]);

  for(i = 0; i < 3; i++){
      printf("Enter first value to find\n");
      scanf("%d", &search);

      first = 0;
      last = n - 1;
      middle = (first+last)/2;

      while (first <= last) {
        if (array[middle] < search)
          first = middle + 1;
        else if (array[middle] == search) {
          printf("%d found at location %d.\n", search, middle+1);
          break;
        }
        else
          last = middle - 1;

        middle = (first + last)/2;
      }
      if (first > last)
        printf("Not found! %d isn't present in the list.\n", search);
   }
   return 0;
}

我如何显示所有 3 个值，如一个低索引、一个高索引和最后一个是最高的，但找不到？

How can i show all the 3 values like one low index, one high and last one is the highest which wont be found?

c

binary-search