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Reverse a singly linked list.
A linked list can be reversed either iteratively or recursively. Could you implement both?
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode reverseList(ListNode head) { if(head == null || head.next == null) { return head; } ListNode p1 = head; ListNode p2 = head.next; head.next = null; while(p1 != null && p2 != null) { ListNode p3 = p2.next; p2.next = p1; p1 = p2; if(p3!= null) { p2 = p3; }else { break; } } return p2; } }
Recursively:
版本一:
public ListNode reverseList(ListNode head) { if(head==null || head.next == null) return head; //get second node ListNode second = head.next; //set first‘s next to be null head.next = null; ListNode rest = reverseList(second); second.next = head; return rest; }
版本二:
public ListNode reverseList(ListNode head) { return reverseListInt(head, null); } public ListNode reverseListInt(ListNode head, ListNode newHead) { if(head == null) return newHead; ListNode next = head.next; head.next = newHead; return reverseListInt(next, head); }
20160601:
iteration:
public class Solution { public ListNode reverseList(ListNode head) { //iteration //base case if(head == null || head.next == null) { return head; } ListNode pre = null; ListNode cur = head; ListNode next = null; while(cur != null) { next = cur.next; cur.next = pre; pre = cur; cur = next; } return pre; } }
recursion:
public class Solution { public ListNode reverseList(ListNode head) { //recursion //base case if(head == null || head.next == null) { return head; } ListNode nextNode = head.next; ListNode newHead = reverseList(nextNode); nextNode.next = head; head.next = null; return newHead; } }
Reference:
1. http://www.programcreek.com/2014/05/leetcode-reverse-linked-list-java/
2. https://leetcode.com/discuss/34474/in-place-iterative-and-recursive-java-solution
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原文地址:http://www.cnblogs.com/anne-vista/p/4799763.html