c语言 链式串中子串的替换算法怎么写

Question

若采用单链存储的方式存储串，编写一个算法将串s中的第i个字符到第
j个字符之间的字符（不包括i和j）用t串替换

Answer 1

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

typedef struct Node {
    char data;
    struct Node* next;
} Node, *LinkedList;

// 创建一个空链表
LinkedList createList() {
    LinkedList L = (LinkedList)malloc(sizeof(Node));
    L->next = NULL;
    return L;
}

// 将一个字符串转换为单链表
LinkedList stringToList(char s[]) {
    LinkedList L = createList();
    Node* tail = L;
    for (int i = 0; s[i] != '\0'; i++) {
        Node* p = (Node*)malloc(sizeof(Node));
        p->data = s[i];
        p->next = NULL;
        tail->next = p;
        tail = p;
    }
    return L;
}

// 将链表转换为一个字符串
void listToString(LinkedList L, char s[]) {
    int i = 0;
    Node* p = L->next;
    while (p != NULL) {
        s[i++] = p->data;
        p = p->next;
    }
    s[i] = '\0';
}

// 将s串中第i个字符到第j个字符之间的字符（不包括i和j）用t串替换
void replaceSubString(LinkedList L, int i, int j, char t[]) {
    // 找到第i个节点前面的节点
    Node* p = L;
    for (int k = 1; k < i; k++) {
        if (p == NULL) {
            printf("i太大\n");
            return;
        }
        p = p->next;
    }
    // 找到第j个节点
    Node* q = p;
    for (int k = i; k <= j; k++) {
        if (q == NULL) {
            printf("j太大\n");
            return;
        }
        q = q->next;
    }
    // 将[t]插入到p和q之间
    Node* tList = stringToList(t);
    Node* tail = tList;
    while (tail->next != NULL) {
        tail = tail->next;
    }
    tail->next = q;
    p->next = tList->next;
    free(tList);
}

int main() {
    char s[] = "Hello, world!";
    LinkedList L = stringToList(s);

    printf("原始串：\n");
    printf("%s\n", s);

    replaceSubString(L, 7, 12, "GitHub");
    char s2[20];
    listToString(L, s2);

    printf("替换后的串：\n");
    printf("%s\n", s2);

    return 0;
}