Question

我正在尝试制作一个程序来计算字符串中每个字符的出现次数，如下所示：

Enter string: hello world

输出为

h: 1
e: 1
l: 3
o: 2

......等等

但我的输出似乎到处都是，例如：

Enter string: nice on ice
n: 2
i: 2
c: 1
e: 1 //there should be 2 e's
o: 1

有时喜欢：

Enter string: potato
p: 1
o: 2
t: 2
a: 1

: 1

这是我的代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

int main() {
  char l, str[30], cpy[30];
  int i, j, occ;
  i = j = 0;
  occ = 0;
  cout << "Enter string: ";
  fgets(str, 10, stdin);
  strcpy(cpy, str); // copy the string as I am about to modify it

  // Checks for occurences of each character.
  // Every character found is changed to a ' ' so that it isn't counted again
  while (str[i] != '\0') {

    occ = 0;
    if (str[i] != ' ') {

      l = str[i];
      for (j = 0; str[j] != '\0'; j++) {

        if (str[j] == cpy[i] && str[j] != ' ') {
          occ++;
          str[j] = ' ';
        }

      }
      cout << l << ": " << occ <<endl;
    }
    i++;
  }

  cout << "\n";
  fputs(str, stdout); // just to check if all characters were converted to ' '
  cout << "\n";
  return 0;
}

Answer 1

你真的不需要复制，你的内循环做了太多的工作，迭代已经失效的字符。另外，您似乎对于i或j是否将副本或原始内容编入索引感到困惑。

int main()
{
    char str[30];
    int i, j, occ;

    cout << "Enter string: ";
    fgets(str, 10, stdin);

    // Checks for occurences of each character.
    // Every character found is changed to a ' ' so that it isn't counted again
    for(i = 0; str[i] != '\0'; i++)
    {
        if (str[i] != ' ')
        {
            // Output the current character before it is invalidated.
            cout << str[i] << ": ";

            // Start at the current character so that it is counted
            // at least once.
            occ = 0;
            for (j = i; str[j] != '\0'; j++)
            {
                // No need to test for ' ', because str[i] is not
                // a ' ', so equality is sufficient.
                if (str[j] == str[i])
                {
                    occ++;
                    // Invalidate each counted character.
                    str[j] = ' ';
                }
            }
            cout << occ << endl;
        }
    }

    // just to check if all characters were converted to ' '
    cout << endl << str << endl;
    return 0;
}

Answer 2

对于您的作业的可能解决方案，请记住英文字母中有26个字母，将每个字母转换为数组的索引很简单（ASCII encoding它是非常简单。）

有了这些信息，请考虑使用26个整数数组，其中数组中的每个元素都是您读取的字符的计数器。

因此，只需在循环中一次读取一个字符，从字符中获取反数组索引，并增加正确的元素。无需实际存储您阅读的文本，无需多次检查循环。

打印时，循环遍历数组并打印每个非零元素及其相应的字符。

检查字符串中每个字符出现的问题

2 个答案: