LeetCode 206. Reverse Linked List

Reverse a singly linked list.

题目大意：反转一个单链表～

分析：设立三个指针：cur——当前结点；pre——当前结点的前一个结点；temp——临时结点（标记cur的next）。首先保存cur的next到temp，然后将cur的next指向pre，将pre移动到当前cur，然后将cur指向temp，直到cur==NULL，返回pre即反转了该链表～

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

class Solution {

public:

ListNode* reverseList(ListNode* head) {

if (head == NULL) return head;

ListNode *cur = head, *pre = NULL, *temp = NULL;

while (cur != NULL) {

temp = cur->next;

cur->next = pre;

pre = cur;

cur = temp;

}

return pre;

}

};

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