Reverse Linked List+链表复习

<pre name="code" class="html">struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
int main(){
ListNode* head=new ListNode(0); //创建第一个链表节点
<span style="white-space:pre">	</span>ListNode* key = head;  //标记链表头的位置
<span style="white-space:pre">	</span>int i;
<span style="white-space:pre">	</span>for (i = 1; i < 5; i++){
<span style="white-space:pre">		</span>ListNode* key = new ListNode(i);//创建链表节点
<span style="white-space:pre">		</span>head->next = key;
<span style="white-space:pre">		</span>head = key;
<span style="white-space:pre">	</span>}
}

while (head!= NULL){
cout << head->val << endl;  //输出链表内容
head = head->next;
}

class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode *key1;
ListNode *key2;
ListNode *key3;
if (head == NULL||head->next==NULL){
return head;
}
key1= head;
key3 = head;
while ((head->next)!= NULL){ //key2指向head当前地址，head指向其下一个地址。通过key2将当前节点指向上一个节点（key1指向）
key2 = head;
head = head->next;
key2->next = key1;
key1 = key2;
}
head->next = key1;//容易遗忘，因为循环条件，必须单独让最后的head指向上一个节点
key3->next = NULL;//同上，key3指向最初的head初始位置。反向后下一个地址指向NULL
return head;
}
};