面试手撕代码——链表反转（递归、迭代）

递归

class Solution {
public:
    ListNode* ReverseList(ListNode* head) {
        if(!head||!head->next) return head;
        ListNode* ans = ReverseList(head->next);
        head->next->next  = head;
        head->next = NULL;
        return ans;
    }
};

迭代

class Solution {
public:
    ListNode* ReverseList(ListNode* pHead) {
        if(!pHead) return NULL;
        ListNode* pre = NULL,*cur = pHead,*nex = NULL;
        while(cur){
            nex = cur->next;
            cur->next = pre;
            pre = cur;
            cur = nex;
        }
        return pre;
    }
};