输入一个链表的头节点,从尾到头反过来返回每个节点的值(用数组返回)。
示例 1:
输入:head = [1,3,2]
输出:[2,3,1]
通常,这种情况下,我们不希望修改原链表的结构。返回一个反序的链表,这就是经典的“后进先出”,我们可以使用栈实现这种顺序。每经过一个结点的时候,把该结点放到一个栈中。当遍历完整个链表后,再从栈顶开始逐个输出结点的值,给一个新的链表结构,这样链表就实现了反转。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
vector<int> reversePrint(ListNode* head) {
stack<int> result;
vector<int> reverse;
ListNode* nodes = head;
while(nodes !=NULL){
result.push(nodes->val);
nodes = nodes->next;
}
while(!result.empty()){
reverse.push_back(result.top());
result.pop();
}
return reverse;
}
};
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def printListFromTailToHead(self, listNode):
# write code here
result = []
while listNode:
result.insert(0, listNode.val)
listNode = listNode.next
return result
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
/**
* 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) {
ListNode* pReversedHead = NULL;
ListNode* pNode = head;
ListNode* pPrev = NULL;
while(pNode != NULL){
ListNode* pNext = pNode->next;
if(pNext == NULL){
pReversedHead = pNode;
}
pNode->next = pPrev;
pPrev = pNode;
pNode = pNext;
}
return pReversedHead;
}
};
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
if not head or not head.next:
return head
last = None
while head:
tmp = head.next
head.next = last
last = head
head = tmp
return last
请判断一个链表是否为回文链表。
示例 1:
输入: 1->2 输出: false
示例 2:
输入: 1->2->2->1 输出: true
在编码的过程中,注意我们比较的是节点值的大小,而不是节点本身。正确的比较方式是:node_1.val == node_2.val。而 node_1 == node_2 是错误的。
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def isPalindrome(self, head: ListNode) -> bool:
vals = []
current_node = head
while current_node is not None:
vals.append(current_node.val)
current_node = current_node.next
return vals == vals[::-1]
输入两个单调递增的链表,输出两个链表合成后的链表,当然我们需要合成后的链表满足单调不减规则。
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
//判断指针是否为空
if(pHead1 == NULL){
return pHead2;
}
else if(pHead2 == NULL){
return pHead1;
}
ListNode* pMergedHead = NULL;
if(pHead1->val < pHead2->val){
pMergedHead = pHead1;
pMergedHead->next = Merge(pHead1->next, pHead2);
}
else{
pMergedHead = pHead2;
pMergedHead->next = Merge(pHead1, pHead2->next);
}
return pMergedHead;
}
};
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
# 返回合并后列表
def Merge(self, pHead1, pHead2):
# write code here
if not pHead1:
return pHead2
if not pHead2:
return pHead1
pMergeHead = None
if pHead1.val < pHead2.val:
pMergeHead = pHead1
pMergeHead.next = self.Merge(pHead1.next, pHead2)
else:
pMergeHead = pHead2
pMergeHead.next = self.Merge(pHead1, pHead2.next)
return pMergeHead