c++ - 从 char 数组中删除重复的单词(仅在后面)

Question

我有点卡住了，无法找出这里出了什么问题。我有一个任务是将一个句子输入到 char 数组中，如果有重复的单词和后面的单词(example :相同的相同，diff diff。但不是:相同的单词相同。)他们应该被删除。这是我写的函数:

void Same(char arr[], char temp[]){
    int i = 0, j = 0, f = 0, *p, k = 0, counter = 0;
    for (i = 0; i < strlen(arr); i++){
        while (arr[i] != ' ' && i < strlen(arr)){
            temp[k] = arr[i];
            i++;
            k++;
            counter++;
        }
        temp[k] = '\0';
        k = 0;
        p = strstr((arr + i), (temp + j));
        if (p != NULL && (*p == arr[i])){
            for (f = 0; f < strlen(p); f++){
                *p = '*';
                p++;
            }
            f = 0;
        }
        j = counter;

    }
}

Answer 1

strtok 是一个方便的函数，可以从列表中获取下一个单词(strsep 是一个更好的函数，但不太可能在您的系统上可用)。使用 strtok，像下面这样的方法可能会起作用，至少对于简单的例子...

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

#define MAXPHRASELEN 1000
#define MAXTOKLEN 100

int main(int argc, char ** argv) 
{
    // Here is the sentence we are looking at
    char * tmp = "This is a test and and another test";
    // We will copy it to this variable
    char phrase[MAXPHRASELEN+1];
    strcpy(phrase, tmp);
    // And will put the altered text in this variable
    char new_phrase[MAXPHRASELEN+1];
    // This will be the last word we looked at
    char * lasttok = malloc(MAXTOKLEN+1);
    // This will be the current word
    char * tok = malloc(MAXTOKLEN+1);
    // Both words are initially empty
    new_phrase[0] = '\0';
    lasttok[0] = '\0';
    // Get the first word
    lasttok = strtok(phrase, " ");
    // If there is a word...
    if (lasttok != NULL) {
        // Put it in the altered text and add a space 
        strcat(new_phrase, lasttok);
        strcat(new_phrase, " ");
        // As long as there is a next word
        while ( (tok = strtok(NULL, " ")) != NULL ) {
            // See if it is the same as the last word
            if (strcmp(tok,lasttok) != 0) {
                // If it isn't, copy it to the altered text
                strcat(new_phrase, tok);
                // and add a space
                strcat(new_phrase, " ");
                // The current word becomes the last word
                lasttok = tok;
            }
        }
    }
    // Print the lot
    printf("%s\n", new_phrase);
}

如果您真的必须编写自己的例程来抓取单个单词，那么您可能比模仿 strtok 做得更糟。它维护一个指向字符串中当前单词开头的指针，并在下一个分隔符(空格字符)处放置一个空字符。再次调用时，它只是将指向字符的指针移到 null 之后，并在下一个分隔符之后放置另一个 null。大多数字符串函数在传递指针时会将 null 视为字符串的结尾，因此只处理当前单词。

减去注释、标题和初始化，它看起来不那么具有威胁性......

    lasttok = strtok(phrase, " ");
    if (lasttok != NULL) {
        strcat(new_phrase, lasttok);
        strcat(new_phrase, " ");
        while ( (tok = strtok(NULL, " ")) != NULL ) {
            if (strcmp(tok,lasttok) != 0) {
                strcat(new_phrase, tok);
                strcat(new_phrase, " ");
                lasttok = tok;
            }
        }
    }
    printf("%s\n", new_phrase);