LeetCode 206. Reverse Linked List 反转单链表

1.题目

反转单链表

2.思路

1) nex 保存当前节点cur的下一个节点
2) cur 指向 前一个节点pre
3) pre = cur 
   cur = nex

 

3.实现

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(head == nullptr)
            return nullptr;
        
        ListNode* pre = nullptr;
        ListNode* cur = head;
        ListNode* nex = nullptr;
        
        while(cur != nullptr)
        {
            nex = cur->next;
            cur->next = pre;
            pre = cur;
            cur = nex;
        }
        
        return pre;
    }
};

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转载自blog.csdn.net/u014128608/article/details/92846700