206. Reverse Linked Listeasy
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206. Reverse Linked List【easy】
Reverse a singly linked list.
Hint:
A linked list can be reversed either iteratively or recursively. Could you implement both?
解法一:
1 class Solution { 2 public: 3 ListNode* reverseList(ListNode* head) { 4 ListNode * pre = NULL; 5 6 while (head != NULL) { 7 ListNode * next = head->next; 8 head->next = pre; 9 pre = head; 10 head = next; 11 } 12 13 return pre; 14 } 15 };
解法二:
1 class Solution { 2 public: 3 ListNode* reverseList(ListNode* head) { 4 if (!head || !(head -> next)) return head; 5 ListNode* node = reverseList(head -> next); 6 head -> next -> next = head; 7 head -> next = NULL; 8 return node; 9 } 10 };
参考了@jianchao.li.fighter 的代码
The basic idea of this recursive solution is to reverse all the following nodes after head
. Then we need to set head
to be the final node in the reversed list. We simply set its next node in the original list (head -> next
) to point to it and sets its next
to be NULL
.
解法三:
1 public ListNode reverseList(ListNode head) { 2 /* recursive solution */ 3 return reverseListInt(head, null); 4 } 5 6 private ListNode reverseListInt(ListNode head, ListNode newHead) { 7 if (head == null) 8 return newHead; 9 ListNode next = head.next; 10 head.next = newHead; 11 return reverseListInt(next, head); 12 }
参考了@braydenCN 的代码
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