如何用c中字串中第二常見的字符替換字串中最常見的字符？

我撰寫了一個 C 函式，它接收來自用戶的字串，并將頻率最高的字母替換為字串中頻率第二高的字母。示例：對于輸入，將回傳i love you more字串i levo yeu mero

#include <stdio.h>
#include <string.h>

void stringReplace(char str1[]);

int main()
{
    char str1[100] = { 0 };
    stringReplace(str1);
    return 0;
}

void stringReplace(char str1[])
{
    char ch1, ch2;
    int i, h, j, p, n, len, counter1 = 0, counter2 = 0, first, second, times;
    printf("Please enter the string - maximum = 100 characters:\n");
    printf("User input: ");
    fgets(str1, 100, stdin);
    str1[strcspn(str1, "\n")] = 0;
    len = strlen(str1);
    for (i = 0; i < len; i  ) {
        counter1 = 0;
        for (j = 0; j < len; j  ) {
            if (str1[i] == str1[j]) {
                counter1  ;
            }
            if (counter1 > counter2) {
                first = i;
            }
        }
        counter2 = counter1;
    } //character which shows up most - found.

    counter2 = 0;

    for (p = 0; p < len; p  ) {
        for (n = 0; n < len; n  ) {
            if (str1[p] == str1[n]) {
                counter1  ;
            }
            if (counter1 < first && counter1 > counter2) {
                second = p;
            }
        }
        counter2 = counter1;
    }

    ch1 = str1[first];
    ch2 = str1[second];
    
    for (h = 0; h < len; h  ) {
        if (str1[h] == ch1) {
            str1[h] = ch2;
        }
    }
    puts(str1);
}

這是我所做的代碼，而不是更改字串它列印相同的東西。例如：

Please enter the string - maximum = 100 characters:
User input: i love you more
i love you more

uj5u.com熱心網友回復：

建議：使用輔助功能分而治之。

想想子問題：計算第 1 和第 2 最受歡迎。

走繩子。在每個字符處計算其與字串其余部分的出現次數。O(n*n)

當檢測到優于第 2 位的計數時，調整第 1 位和第 2 位。

#include <ctype.h>
#include <stdio.h>
#include <stdlib.h>

void most_popular2(const char *s, char pop[2]) {
  pop[0] = pop[1] = 0;
  size_t pop_count[2] = {0, 0};

  while (*s) {
    // Do not count white-space.
    if (!isspace(*(const unsigned char* )s)) {
      // How popular is *s?
      size_t count = 1;
      // sum all the matches on the rest of the string
      for (const char *r = s   1; *r; r  ) {
        if (*s == *r) {
          count  ;
        }
      }

      // Test if better than the 2nd most popular.
      if (count > pop_count[1] && *s != pop[0]) {
        if (count > pop_count[0]) {
          // Demote 1st place to 2nd place.
          pop_count[1] = pop_count[0];
          pop[1] = pop[0];
          // Save new 1st place.
          pop_count[0] = count;
          pop[0] = *s;
        } else {
          // Save new 2nd place.
          pop_count[1] = count;
          pop[1] = *s;
        }

      }
    }
    s  ;
  }
}

樣本

int main() {
  char s[] = "i love you more i love you more\n";
  char pop[2];
  most_popular2(s, pop);
  printf("1st:%c 2nd:%c\n", pop[0], pop[1]);
}

輸出

1st:o 2nd:e

將其留給 OP 以獲取 2 個最受歡迎的字符并形成所需的輸出。

更高效的代碼可以復制字串，對它進行 O(n*(lng(n)) 排序，然后遍歷它以找到最流行的 2 個。

uj5u.com熱心網友回復：

謝謝你的幫助，現在我明白了。這就是我所做的：

#include <stdio.h>
#include<stdio.h> 
#include<string.h> 

#define ASCII_A 97  // ascii for A is 97 
#define MAXSIZE 50 // max size for string 
#define ALPHABETS 26 //total number of alphabets 

void stringReplace(char s[], int a[]);

int main()
{
    char s[MAXSIZE] = {0};
    int a[ALPHABETS] = {0}
    stringReplace(s);
    return 0;
}
/*
function to replace the most common charcter with the second one
input: the string
output: none
*/
void stringReplace(char s[], int a[])
{
    int i;
    int count = 0;
    int max = 0, max2 = 0;
    int index, index2; 

    printf("Enter the string:\n"); 
    fgets(s, MAXSIZE, stdin);
    s[strcspn(s, "\n")] = 0;

    for(i=0; i<26; i  ) // initialize all array elements to zero 
    {
        a[i]=0;
    }       

    for(i=0; i<strlen(s); i  ) // loop for traversing the string 
    {
        a[s[i]-ASCII_A]  ; // increment by 1 for each occurrence. 
    }

    for(i=0; i< ALPHABETS; i  ) //prints alphabets and occurrence. 
    {
        count = a[i];
        if (count > max) 
        {
            max = count;
            index = i;
        }
        if(count < max && count > 1)
        {
            max2 = count;
            index2 = i;
        }   
    }
    printf("%c\n", ASCII_A index);
    printf("%c\n", ASCII_A index2);

}

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