# Reverse Linked List ### Source - leetcode: [Reverse Linked List | LeetCode OJ](https://leetcode.com/problems/reverse-linked-list/) - lintcode: [(35) Reverse Linked List](http://www.lintcode.com/en/problem/reverse-linked-list/) ~~~ Reverse a linked list. Example For linked list 1->2->3, the reversed linked list is 3->2->1 Challenge Reverse it in-place and in one-pass ~~~ ### 题解1 - 非递归 联想到同样也可能需要翻转的数组，在数组中由于可以利用下标随机访问，翻转时使用下标即可完成。而在单向链表中，仅仅只知道头节点，而且只能单向往前走，故需另寻出路。分析由`1->2->3`变为`3->2->1`的过程，由于是单向链表，故只能由1开始遍历，1和2最开始的位置是`1->2`，最后变为`2->1`，故从这里开始寻找突破口，探讨如何交换1和2的节点。 ~~~ temp = head->next; head->next = prev; prev = head; head = temp; ~~~ 要点在于维护两个指针变量`prev`和`head`, 翻转相邻两个节点之前保存下一节点的值，分析如下图所示：  1. 保存head下一节点 1. 将head所指向的下一节点改为prev 1. 将prev替换为head，波浪式前进 1. 将第一步保存的下一节点替换为head，用于下一次循环 ### Python ~~~ # Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: # @param {ListNode} head # @return {ListNode} def reverseList(self, head): prev = None curr = head while curr is not None: temp = curr.next curr.next = prev prev = curr curr = temp # fix head head = prev return head ~~~ ### C++ ~~~ /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* reverse(ListNode* head) { ListNode *prev = NULL; ListNode *curr = head; while (curr != NULL) { ListNode *temp = curr->next; curr->next = prev; prev = curr; curr = temp; } // fix head head = prev; return head; } }; ~~~ ### Java ~~~ /** * 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) { ListNode prev = null; ListNode curr = head; while (curr != null) { ListNode temp = curr.next; curr.next = prev; prev = curr; curr = temp; } // fix head head = prev; return head; } } ~~~ ### 源码分析 题解中基本分析完毕，代码中的prev赋值比较精炼，值得借鉴。 ### 复杂度分析 遍历一次链表，时间复杂度为 O(n)O(n)O(n), 使用了辅助变量，空间复杂度 O(1)O(1)O(1). ### 题解2 - 递归 递归的终止步分三种情况讨论： 1. 原链表为空，直接返回空链表即可。 1. 原链表仅有一个元素，返回该元素。 1. 原链表有两个以上元素，由于是单链表，故翻转需要自尾部向首部逆推。 由尾部向首部逆推时大致步骤为先翻转当前节点和下一节点，然后将当前节点指向的下一节点置空(否则会出现死循环和新生成的链表尾节点不指向空)，如此递归到头节点为止。新链表的头节点在整个递归过程中一直没有变化，逐层向上返回。 ### Python ~~~ """ Definition of ListNode class ListNode(object): def __init__(self, val, next=None): self.val = val self.next = next """ class Solution: """ @param head: The first node of the linked list. @return: You should return the head of the reversed linked list. Reverse it in-place. """ def reverse(self, head): # case1: empty list if head is None: return head # case2: only one element list if head.next is None: return head # case3: reverse from the rest after head newHead = self.reverse(head.next) # reverse between head and head->next head.next.next = head # unlink list from the rest head.next = None return newHead ~~~ ### C++ ~~~ /** * Definition of ListNode * * class ListNode { * public: * int val; * ListNode *next; * * ListNode(int val) { * this->val = val; * this->next = NULL; * } * } */ class Solution { public: /** * @param head: The first node of linked list. * @return: The new head of reversed linked list. */ ListNode *reverse(ListNode *head) { // case1: empty list if (head == NULL) return head; // case2: only one element list if (head->next == NULL) return head; // case3: reverse from the rest after head ListNode *newHead = reverse(head->next); // reverse between head and head->next head->next->next = head; // unlink list from the rest head->next = NULL; return newHead; } }; ~~~ ### Java ~~~ /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode reverse(ListNode head) { // case1: empty list if (head == null) return head; // case2: only one element list if (head.next == null) return head; // case3: reverse from the rest after head ListNode newHead = reverse(head.next); // reverse between head and head->next head.next.next = head; // unlink list from the rest head.next = null; return newHead; } } ~~~ ### 源码分析 case1 和 case2 可以合在一起考虑，case3 返回的为新链表的头节点，整个递归过程中保持不变。 ### 复杂度分析 递归嵌套层数为 O(n)O(n)O(n), 时间复杂度为 O(n)O(n)O(n), 空间(不含栈空间)复杂度为 O(1)O(1)O(1). ### Reference - [全面分析再动手的习惯：链表的反转问题（递归和非递归方式） - 木棉和木槿 - 博客园](http://www.cnblogs.com/kubixuesheng/p/4394509.html) - [data structures - Reversing a linked list in Java, recursively - Stack Overflow](http://stackoverflow.com/questions/354875/reversing-a-linked-list-in-java-recursively) - [反转单向链表的四种实现（递归与非递归，C++） | 宁心勉学，慎思笃行](http://ceeji.net/blog/reserve-linked-list-cpp/) - [iteratively and recursively Java Solution - Leetcode Discuss](https://leetcode.com/discuss/37804/iteratively-and-recursively-java-solution)