[难度] easy
[类别] linked list
1.题目描述
反转链表
2.算法描述
(1)第一种算法实现
使用栈作为中间辅助。
首先将链表的所有节点push到栈中,然后再从栈中pop出来,即可得到反转链表
代码实现:
ListNode* reverseList(ListNode* head) {
stack<ListNode*> mystack;
// get reversed list l1
ListNode* p = head;
while (p != NULL) {
mystack.push(p);
p = p->next;
}
if (!mystack.empty()) {
p = mystack.top();
mystack.pop();
head = p;
}
while (!mystack.empty()) {
p->next = mystack.top();
mystack.pop();
p = p->next;
}
if (p != NULL)
p->next = NULL;
return head;
}
(2)第二种算法
不另外开辟节点空间,实现链表的反转
ListNode* reverseList(ListNode* head) {
if (head == NULL || head->next == NULL) return head;
ListNode* result = head;
ListNode* temp = head->next;
head = temp->next;
result->next = NULL;
while (head != NULL) {
temp->next = result;
result = temp;
temp = head;
head = head->next;
}
temp->next = result;
result = temp;
return result;
}
测试代码:
void print(ListNode* l1) {
cout << l1->val;
l1 = l1->next;
int count = ;
while (l1 != NULL) {
cout << " -> " << l1->val;
l1 = l1->next;
count++;
}
cout << " length: " << count << endl;
}
int main() {
int n1;
cout << "len1:"
cin >> n1
int num;
ListNode* head = NULL;
ListNode* l1 = NULL;
int i = n1;
while (i--) {
cin >> num;
if (i == (n1 -)) {
head = new ListNode(num);
l1 = head;
}
else {
head->next = new ListNode(num);
head = head->next;
}
}
print(l1);
print(reverseList(l1));
system("pause");
return ;
}
3.小结
第一种方法比较简单,直接使用栈作为中间过渡;第二种方法逻辑性强,代码简洁且不需要另外开辟内存空间。第二种方法的实现比较容易出错,要特别注意在最开始的时候设置result->next = NULL。