35. 翻转链表(reverse-linked-list)(c++)----lintcode面试题之链表
2018-02-16 21:14
543 查看
(一)题目要求:翻转一个链表(二)示例:给出一个链表1->2->3->null,这个翻转后的链表为3->2->1->null(三)题解:方法一:/**
* Definition of ListNode
*
* class ListNode {
* public:
* int val;
* ListNode *next;
*
* ListNode(int val) {
* this->val = val;
* this->next = NULL;
* }
* }
*/
//方法一:只有一个结点时无操作
//p_front 指向头结点 p_insert指向头结点的下一个结点
//下一个结点更新
//p_insert插入链表头
//p_insert置为下一个结点
class Solution {
public:
/*
* @param head: n
* @return: The new head of reversed linked list.
*/
ListNode * reverse(ListNode * head) {
// write your code here
ListNode *p_front = head,*p_insert = NULL;
while(p_front && (p_insert = p_front->next))
{
p_front->next = p_insert->next;
p_insert->next = head;
head = p_insert;
}
return head;
}
};方法二://方法二
//和方法一基本相同 不同是 先将prev置为NULL,使第一个结点也参与逆序操作
class Solution {
public:
/**
* @param head: The first node of linked list.
* @return: The new head of reversed linked list.
*/
ListNode *reverse(ListNode *head) {
ListNode *prev = NULL;
while (head != NULL) {
ListNode *temp = head->next;
head->next = prev;
prev = head;
head = temp;
}
return prev;
}
};
* Definition of ListNode
*
* class ListNode {
* public:
* int val;
* ListNode *next;
*
* ListNode(int val) {
* this->val = val;
* this->next = NULL;
* }
* }
*/
//方法一:只有一个结点时无操作
//p_front 指向头结点 p_insert指向头结点的下一个结点
//下一个结点更新
//p_insert插入链表头
//p_insert置为下一个结点
class Solution {
public:
/*
* @param head: n
* @return: The new head of reversed linked list.
*/
ListNode * reverse(ListNode * head) {
// write your code here
ListNode *p_front = head,*p_insert = NULL;
while(p_front && (p_insert = p_front->next))
{
p_front->next = p_insert->next;
p_insert->next = head;
head = p_insert;
}
return head;
}
};方法二://方法二
//和方法一基本相同 不同是 先将prev置为NULL,使第一个结点也参与逆序操作
class Solution {
public:
/**
* @param head: The first node of linked list.
* @return: The new head of reversed linked list.
*/
ListNode *reverse(ListNode *head) {
ListNode *prev = NULL;
while (head != NULL) {
ListNode *temp = head->next;
head->next = prev;
prev = head;
head = temp;
}
return prev;
}
};
相关文章推荐
- [Lintcode]Reverse Linked List II 翻转链表 II
- lintcode 中等题: reverse linked list II 翻转链表II
- 466. 链表节点计数 (count-linked-list-nodes)(c++)----lintcode面试题之链表
- 372. 在O(1)时间复杂度删除链表节点 (delete-node-in-the-middle-of-singly-linked-list)(c++)----lintcode面试题之链表
- 【LintCode】 Reverse Linked List 翻转链表
- Reverse Linked List II 局部翻转链表@LeetCode
- Reverse Linked List II -- 翻转部分链表
- 【LeetCode】 Reverse Linked List 翻转链表 ( Uber,FB ) - Medium++
- leetcode之92. Reverse Linked List II(C++读错题版本,交换一个链表中指定的两个位置上的元素)
- 部分翻转链表Reverse Linked List II
- 剑指offer 面试题16:反转链表(Leetcode 206: Reverse Linked List) 题解
- [LeetCode] Reverse Linked List I II - 链表翻转问题
- LeetCode Reverse Linked List翻转链表
- Reverse Linked List II 翻转链表II
- LeetCode-Reverse Linked List II-翻转链表区间-链表操作
- 174. 删除链表中倒数第n个节点 (remove-nth-node-from-end-of-list)(c++)----lintcode面试题之链表
- 166. 链表倒数第n个节点(nth-to-last-node-in-list)(c++)----lintcode面试题之链表
- lintcode 容易题:Reverse Linked List 翻转链表
- LeetCode 92. Reverse Linked List II(翻转链表)
- 173. 链表插入排序 (insertion-sort-list)(c++)----lintcode面试题之链表