题解 | #反转链表#
反转链表
https://www.nowcoder.com/practice/75e878df47f24fdc9dc3e400ec6058ca
/** * struct ListNode { * int val; * struct ListNode *next; * ListNode(int x) : val(x), next(nullptr) {} * }; */ class Solution { public: /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param head ListNode类 * @return ListNode类 */ ListNode* ReverseList(ListNode* head) { // write code here if ( head == nullptr || head->next == nullptr){ return head; } ListNode * prev = head; ListNode *curr = prev->next; prev-> next = nullptr; // p0 c1 t2 while(curr!=nullptr){ ListNode* next = curr->next; // cout<<to->val; curr->next = prev; prev = curr; curr = next; } return prev; } };
⚠️: 这是模板,注意while边界条件,返回值是prev