本来没想出来，刚才突然想到，可以用“头插法”来反转
Reverse Linked List
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Reverse a singly linked list.
Discuss中的递归解法：
1 public ListNode reverseList(ListNode head) { 2 return reverseListInt(head, null); 3 } 4 5 public ListNode reverseListInt(ListNode head, ListNode newHead) { 6 if(head == null) 7 return newHead; 8 ListNode next = head.next; 9 head.next = newHead; 10 return reverseListInt(next, head); 11 }
我的代码：用的是迭代（即循环）的方法
C语言：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * struct ListNode *next; 6 * }; 7 */ 8 struct ListNode* reverseList(struct ListNode* head) { 9 if (head == NULL || head->next == NULL) 10 return head; 11 struct ListNode* p = head->next; 12 struct ListNode* q = p; 13 head->next = NULL; 14 while(p) { 15 p = p->next; 16 q->next = head; 17 head = q; 18 q = p; 19 } 20 21 return head; 22 }