leetcode 206. Reverse Linked List(剑指offer16)、

206. Reverse Linked List

之前在牛客上的写法：

错误代码:

class Solution {
public:
    ListNode* ReverseList(ListNode* pHead) {
        if(pHead == NULL)
            return NULL;
        ListNode* p1 = pHead;
        ListNode* p2 = pHead->next;
        ListNode* p3 = pHead->next->next;
        pHead->next = NULL;
        while(p3 != NULL){
            p2->next = p1;
            p1 = p2;
            p2 = p3;
            p3 = p3->next;
        }
        p2->next = p1;
        return p2;
    }
};

此代码会报“段错误:您的程序发生段错误，可能是数组越界，堆栈溢出（比如，递归调用层数太多）等情况引起”

如果链表只有一个节点，那p2就是空指针，p3就是空指针的下一个指针，但空指针是没有next的

正确代码:

class Solution {
public:
    ListNode* ReverseList(ListNode* pHead) {
        if(pHead == NULL)
            return NULL;
        else if(pHead->next == NULL)
            return pHead;
        ListNode* p1 = pHead;
        ListNode* p2 = pHead->next;
        ListNode* p3 = pHead->next->next;
        pHead->next = NULL;
        while(p3 != NULL){
            p2->next = p1;
            p1 = p2;
            p2 = p3;
            p3 = p3->next;
        }
        p2->next = p1;
        return p2;
    }
};

个人觉得这种以当前节点为循环判断条件的方式比较好：

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* pre = NULL;
        while(head){
            ListNode* tmp = head->next;
            head->next = pre;
            pre = head;
            head = tmp;
        }
        return pre;
    }
};