反转链表的解法分享

1、双指针法

 ListNode* addInList(ListNode* head1, ListNode* head2) {
        // write code here
       ListNode* ReverseList(ListNode* pHead){
            if(pHead == NULL)
                return NULL;
            ListNode* cur = pHead;
            ListNode* pre =NULL;
            while(cur != NULL)
            {
                ListNode* temp = cur->next;
                cur->next = pre;
                pre = cur;
                cur = temp;
            }
            return  pre;
       } 
    }

2、递归写法

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        return reverse(head,nullptr);
    }
    ListNode* reverse(ListNode* cur, ListNode* pre) {
        if (cur == nullptr) return pre;

        ListNode* temp = cur->next;
        cur->next = pre;
        return reverse(temp, cur); // 添加return语句以返回新链表的头结点
    }
};