两个面试题

1. 实现一个lite版的字符串替换函数

C/C++:

  char *strreplace(char *str, const char *sub,const char *rep)

限制条件和要求如下:

1.其中str为原字符串，sub为待被替换的子串。为简单起见, 假定字符串sub和rep长度一样

2.直接对原字符串str进行修改并返回, 不得使用malloc/new开辟新的内存空间

3.不得使用任何库函数/API, 包括但不限于strlen, strstr, strcpy, 如需使用类似功能, 请自行实现

#include <iostream>#include <cstdlib>using namespace std;//此函数支持原字符串中有多个子串的情况，将分别替换//多个子字符串重叠将处理为替换后跳过已替换部分，如aaabcd, aa, xx运行结果为xxabcd而不是xxxbcdchar *strreplace(char *str, const char *sub, const char *rep){char *ret = str;while (*str != '\0'){int i = 0;for (; *(sub + i) != '\0'; i++){if (*(str + i) != *(sub + i)) break;}if (*(sub + i) == '\0'){for (int j = 0; *(rep + j) != '\0'; j++)*str++ = *(rep + j);}else str++;}return ret;}int main(){char a[] = "fuck you mother fucker";char b[] = "fuck";char c[] = "shit";char *d = strreplace(a, b, c);while (*d != '\0')cout << *d++;cout << endl;system("Pause");return 0;}

2. 假设有两个单链表A和B，不带头节点，且长度一样，示例如下：

A:1->2->3->4

B:a->b->c->d

请逆转交替合并两个链表，示例结果如下：

4->d->3->c->2->b->1->a

 

C/C++:

节点类型定义如下:

structNode {

    struct Node *next;

    ...

}

函数定义如下:

Node*reverse_merge(Node *A, Node *B)

其中A、B分别是指向对应链表的第一个节点的指针

请直接修改原有链表的next指针完成该操作，并返回新链表的第一个节点的指针

#include <iostream>#include <cstdlib>using namespace std;struct Node {struct Node *next;char data;};//C/C++实现Node *reverse_merge(Node *A, Node *B){//前两个if防御式编程if (A == NULL) return A;if (A->next == NULL){B->next = 0;A->next = B;return A;}Node *A1 = A->next;Node *B1 = B->next;Node *B2 = B1->next;B->next = 0;while (A1 != NULL){A->next = B;B1->next = A;A = A1;A1 = A1->next;B = B1;B1 = B2;//运行到修改链表最后节点时B2已为NULL，不可再取其nextif(B2!=NULL) B2 = B2->next;}A->next = B;return A;}Node *buildList(Node *A, char c){Node *ret = A;while (A->next != NULL)A = A->next;Node *p = (Node*)malloc(sizeof(Node));p->data = c;p->next = NULL;A->next = p;return ret;}int main(){Node *A = (Node*)malloc(sizeof(Node));Node *B = (Node*)malloc(sizeof(Node));A->data = '1';B->data = 'a';A->next = B->next = NULL;A = buildList(A, '2');A = buildList(A, '3');A = buildList(A, '4');//A = buildList(A, '5');//A = buildList(A, '6');//分隔符B = buildList(B, 'b');B = buildList(B, 'c');B = buildList(B, 'd');//B = buildList(B, 'e');//B = buildList(B, 'f');/*while (B != NULL){cout << B->data << " ";B = B->next;}*/Node *C = reverse_merge(A, B);while (C != NULL){cout << C->data << " ";C = C->next;}cout << endl;system("Pause");return 0;}