206.反转链表
1.题目描述
反转一个单链表。
示例:
输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL
进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题？
2.解题报告
思路1：借助栈
利用栈先进后出的特点，将每个节点按顺序存入栈中，再从顶到底连接栈中的每个节点
注意要将翻转后的最后一个节点（即原链表的第一个节点）的next置为nullptr，不然后果可想而知~
思路2：就地操作（推荐）
逐个断开原链表的每个节点（保存下个节点）
将断开的节点连接到反转链表的表头上
更新反转链表的表头
回到原链表的下个节点
3.最优答案
c答案
/**
* Definition for singly-linked list.
* struct ListNode {
*
int val;
*
struct ListNode *next;
* };
*/
//struct ListNode* reverseList(struct ListNode* head) {
// struct ListNode* new = NULL;
// while(head) {
//
struct ListNode* temp = head;
//
head = head->next;
//
temp->next = new;
//
new = temp;
// }
// return new;
//}
struct ListNode* reverseList(struct ListNode* head) {
if (!head || !head->next) return head;
struct ListNode *L = reverseList(head->next);
head->next->next = head;
head->next = NULL;
return L;
}
c++答案
/**
* Definition for singly-linked list.
* struct ListNode {
*
int val;
*
ListNode *next;
*
ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if (!head || !head->next) return head;
ListNode *node = reverseList(head->next);
head->next->next = head;
head->next = NULL;
return node;
}
};
java答案
/**
* Definition for singly-linked list.
* public class ListNode {
*
int val;
*
ListNode next;
*
ListNode(int x) { val = x; }
* }
*/
class DIYStack{
public ArrayList<Integer> container = new ArrayList<>();
public void comeIn(int item){
container.add(item);
}
public int comeOut(){
return container.remove(container.size()-1);
}
}
class Solution {
public ListNode reverseList(ListNode head) {
DIYStack diyStack = new DIYStack();
ListNode tmp = head;
while (tmp != null){
diyStack.comeIn(tmp.val);
tmp = tmp.next;
}
tmp = head;
while (tmp != null) {
tmp.val = diyStack.comeOut();
tmp = tmp.next;
}
return head;
}
}
JavaScript答案
/**
* Definition for singly-linked list.
* function ListNode(val) {
*
this.val = val;
*
this.next = null;
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var reverseList = function(head) {
var cur = head;
var prev = null;
while(cur){
var next = cur.next;
cur.next = prev;
prev = cur;
cur = next;
}
return prev;
};
c#答案
/**
* Definition for singly-linked list.
* public class ListNode {
*
public int val;
*
public ListNode next;
*
public ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode ReverseList(ListNode head) {
ListNode b = null;
ListNode Nextindex;
while(head != null)
{
Nextindex = head.next;
head.next = b;
b=head;
head = Nextindex;
}
return b;
}
}
python2.x答案
# Definition for singly-linked list.
# class ListNode(object):
#
def __init__(self, x):
#
self.val = x
#
self.next = None
class Solution(object):
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if head is None or head.next is None:
return head
stack = []
while head:
stack.append(head)
head = head.next
newHead = stack[-1]
# while stack:
#
now = stack.pop()
for i in range(len(stack) - 1, 0, -1):
stack[i].next = stack[i - 1]
stack[0].next = None
return newHead
python3.x答案
# Definition for singly-linked list.
# class ListNode:
#
def __init__(self, x):
#
self.val = x
#
self.next = None
class Solution:
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if head is None:
return None
if head.next is None:
return head
p = head
q = None
while p:
tmp = p.next
p.next = q
q = p
p = tmp
return q
go答案
/**
* Definition for singly-linked list.
* type ListNode struct {
*
Val int
*
Next *ListNode
* }
*/
func reverseList(head *ListNode) *ListNode {
var preNode *ListNode = nil
var currentNode *ListNode = head
for currentNode != nil {
nextNode := currentNode.Next
currentNode.Next = preNode
preNode = currentNode
currentNode = nextNode
}
return preNode
}
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