LeetCode - 206. Reverse Linked List
2016-06-29 21:03
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迭代法:采用三个指针,分别指向前一个结点,当前结点和下一个结点,将当前结点的next指针指向下一个结点,然后更新前一个节点(prev++)和当前结点(curr++),时间复杂度O(n),代码如下:
/**
* 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) {
// Corner case
if(head == null){
return head;
}
ListNode prev = null;
ListNode curr = head;
while(curr != null){
ListNode next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
head = prev;
return head;
}
}
递归:首先获取head.next结点,对这个结点调用reverseList,然后假设rest都已经reverse完毕的情况下调整当前结点的指针,时间复杂度O(n),代码如下:
public class Solution{
public ListNode reverseList(ListNode head){
if(head == null){
return head;
}
ListNode rest = head.next;
reverseList(rest);
head.next.next = head;
head.next = null;
head = rest;
return head;
}
}
知识点:
1. 在Linked List和Array中,采用多个指针是一个很好的思路
2. 这道题目是一个关于Linked List的基本技能题目,多多记下
/**
* 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) {
// Corner case
if(head == null){
return head;
}
ListNode prev = null;
ListNode curr = head;
while(curr != null){
ListNode next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
head = prev;
return head;
}
}
递归:首先获取head.next结点,对这个结点调用reverseList,然后假设rest都已经reverse完毕的情况下调整当前结点的指针,时间复杂度O(n),代码如下:
public class Solution{
public ListNode reverseList(ListNode head){
if(head == null){
return head;
}
ListNode rest = head.next;
reverseList(rest);
head.next.next = head;
head.next = null;
head = rest;
return head;
}
}
知识点:
1. 在Linked List和Array中,采用多个指针是一个很好的思路
2. 这道题目是一个关于Linked List的基本技能题目,多多记下
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