题解 | 反转链表

/**
 * 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 == NULL) {
            return NULL;
        }

        ListNode* prev = NULL;
        ListNode* curr = head; // head is not NULL
        ListNode* next = head;

        while (curr != NULL) {
            next = curr->next;
            curr->next = prev;
            prev = curr;
            curr = next;
            
        }

        return prev;

    }
};

通过辅助的prev，curr和next来记录，做值的inverse然后完成in place的链表掉头#剑指offer#