题目描述
输入一个链表,按链表值从尾到头的顺序返回一个ArrayList。
链表的结构
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};
1.利用栈的特性
class Solution {
public:
vector<int> printListFromTailToHead(ListNode* head)
{
stack<int> s ;
vector<int> v;
if(nullptr == head)
return v;
ListNode* cur = head;
while(cur)
{
s.push(cur->val);
cur = cur -> next;
}
while(!s.empty())
{
v.push_back(s.top());
s.pop();
}
return v;
}
};
2.利用递归
class Solution {
public:
vector<int> printListFromTailToHead(ListNode* head)
{
vector<int> v;
if(nullptr == head)
return v;
v = printListFromTailToHead(head->next);
v.push_back(head->val);
return v;
}
};
3.利用数组
class Solution {
public:
vector<int> printListFromTailToHead(ListNode* head)
{
vector<int> v;
if(head == nullptr)
return v;
ListNode* cur = head;
while(cur)
{
v.push_back(cur->val);
cur = cur->next;
}
// reverse(v.begin(),v.end()); //也可以用STL直接提供的反转函数
int temp=0;
int i=0,j=value.size()-1;
while(i<j)
{
temp=value[i]; //也可以用swap函数,swap(value[i],value[j]);
value[i]=value[j];
value[j]=temp;
i++;
j--;
}
return value;
}
};
4.利用反向迭代器
class Solution {
public:
vector<int> printListFromTailToHead(ListNode* head)
{
vector<int> v;
if(head == nullptr)
return v;
ListNode* cur = head;
while(cur)
{
v.push_back(cur->val);
cur = cur->next;
}
return vector<int>(v.rbegin(), v.rend());
}
};
<stron>:逆序打印单链表</stron>