用另一个给定单词替换文本中的单词的C程序

给定三个字符串“ str”，“ oldW”和“ newW”。该任务是找到单词“ oldW”的所有出现，然后替换为单词“ newW”。
例子：

Input : str[] = "xxforxx xx for xx", 
        oldW[] = "xx", 
        newW[] = "geeks"
Output : geeksforgeeks geeks for geeks

我们的想法是穿越原始字符串并计算在字符串中出现旧词的次数。现在制作一个足够大的新字符串，以便可以替换新单词。现在，用单词替换将原始字符串复制到新字符串。

// C program to search and replace
// all occurrences of a word with
// other word.
#include 
#include 
#include 
  
// Function to replace a string with another
// string
char* replaceWord(const char* s, const char* oldW,
                  const char* newW)
{
    char* result;
    int i, cnt = 0;
    int newWlen = strlen(newW);
    int oldWlen = strlen(oldW);
  
    // Counting the number of times old word
    // occur in the string
    for (i = 0; s[i] != '\0'; i++) {
        if (strstr(&s[i], oldW) == &s[i]) {
            cnt++;
  
            // Jumping to index after the old word.
            i += oldWlen - 1;
        }
    }
  
    // Making new string of enough length
    result = (char*)malloc(i + cnt * (newWlen - oldWlen) + 1);
  
    i = 0;
    while (*s) {
        // compare the substring with the result
        if (strstr(s, oldW) == s) {
            strcpy(&result[i], newW);
            i += newWlen;
            s += oldWlen;
        }
        else
            result[i++] = *s++;
    }
  
    result[i] = '\0';
    return result;
}
  
// Driver Program
int main()
{
    char str[] = "xxforxx xx for xx";
    char c[] = "xx";
    char d[] = "Geeks";
  
    char* result = NULL;
  
    // oldW string
    printf("Old string: %s\n", str);
  
    result = replaceWord(str, c, d);
    printf("New String: %s\n", result);
  
    free(result);
    return 0;
}

输出：

Old string: xxforxx xx for xx
New String: GeeksforGeeks Geeks for Geeks

方法2：此方法涉及到字符串的就地更新。由于它仅使用多余的空间来插入新字符，因此效率更高。

#include 
using namespace std;
int main()
{
    // code
    int t;
    cin >> t;
    cin.ignore();
  
    while (t--) {
        string s;
        getline(cin, s);
  
        string x, y;
        getline(cin, x);
        getline(cin, y);
  
        reverse(s.begin(), s.end());
        reverse(x.begin(), x.end());
        reverse(y.begin(), y.end());
  
        int ls = s.length(), lx = x.length(), ly = y.length();
        int d = ly - lx;
        int ct = 0;
        int i = 0, j = 0;
  
        while (i < ls) {
            string temp = "";
            for (int k = 0; k < lx; k++) {
                temp += s[i + k];
            }
  
            if (temp == x) {
                ct++;
                i = i + lx;
            }
            else {
                i = i + 1;
            }
        }
  
        for (int i = 0; i < ct * d; i++)
            s += ' ';
  
        i = ls - 1;
        j = ls + ct * d - 1;
  
        while (i >= 0 && j >= 0) {
            string temp = "";
            for (int k = 0; k < lx; k++) {
                if (i < (lx - 1 - k))
                    break;
  
                temp += s[i - (lx - 1 - k)];
            }
  
            if (temp == x) {
                int k = ly - 1;
                while (k >= 0)
                    s[j--] = y[k--];
  
                i = i - lx;
            }
            else {
                s[j--] = s[i--];
            }
        }
        reverse(s.begin(), s.end());
        cout << s << endl;
    }
    return 0;
}

输入：

1
xxforxx xx for xx
xx
geeks

输出：

geeksforgeeks geeks for geeks

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