206.反转链表

反转一个单链表。

示例:

输入: 1->2->3->4->5->NULL

输出: 5->4->3->2->1->NULL

 

来源：力扣（LeetCode）

链接：https://leetcode-cn.com/problems/reverse-linked-list

著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

 

 

struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode *new_head = NULL;//指向新链表头节点的指针
        while (head) {
            //备份 head->next,因为第二步要把head->next 指向新的头节点,不保存就找不到了
            ListNode *next = head->next;
            head->next = new_head;//更新 head->next,指向新链表的头节点
            new_head = head;//移动 new_head，指向新节点
            head = next;
        }
        return new_head;
    }
};

 

测试

int main(int argc, const char * argv[]) {
    ListNode a(1);
    ListNode b(2);
    ListNode c(3);
    ListNode d(4);
    ListNode e(5);
    a.next = &b;
    b.next = &c;
    c.next = &d;
    d.next = &e;
    Solution solve;
    ListNode *head = &a;
    cout << "Before
";
    while (head) {
        cout <<head->val<<endl;
        head = head->next;
    }
    head = solve.reverseList(&a);
    cout << "After
";
    while (head) {
        cout <<head->val<<endl;
        head = head->next;
    }
    
    return 0;
}