C++单向链表反转

C++单向链表反转

基本原理

遍历一遍链表，当访问到下一个节点时，将当前指针块与上一块指针块相连。

根据我的理解，我这里用到用到三个指针

假设有如下链表：



新建三个指针，信息分别如下：



*Head、*Curr都指向List，而*Temp指向一个空块

然后将Curr移动到下一块链表，这时将Head->Next指向Temp



最后移动Temp到Head，再将Head移动到Curr，整个过程大概就这样：



代码如下：

Node* My_reverse(Node *head) {
if (head == NULL) {
return head;
}
Node* temp = NULL;
Node* current_node = head;
while(current_node->next != NULL) {
current_node = current_node->next;
head->next = temp;
temp = head;
head = current_node;
}
head->next = temp;
return head;
}

整个测试代码如下：

#include <iostream>
using namespace std;

struct Node {
int num;
Node* next;
Node(int a = 0) {
num = a;
next = NULL;
}
};

Node* My_reverse(Node *head) {
if (head == NULL) {
return head;
}
Node* temp = NULL;
Node* current_node = head;
while(current_node->next != NULL) {
current_node = current_node->next;
head->next = temp;
temp = head;
head = current_node;
}
head->next = temp;
return head;
}

int main() {
Node *temp = new Node();
Node *tail = temp;
for (int i = 1; i < 5; i++) {
Node *new_node = new Node(i);
tail->next = new_node;
tail = new_node;
}
Node *a = temp;
while (a != NULL) {
cout << a->num << ' ';
a = a->next;
}
cout << endl;
a = My_reverse(temp);
while (a != NULL) {
cout << a->num << ' ';
a = a->next;
}
cout << endl;
return 0;
}

输出结果：