Remove Nth Node From End of List
最后更新于：2022-04-02 01:10:26
# Remove Nth Node From End of List
### Source
- lintcode: [(174) Remove Nth Node From End of List](http://www.lintcode.com/en/problem/remove-nth-node-from-end-of-list/)
~~~
Given a linked list, remove the nth node from the end of list and return its head.
Note
The minimum number of nodes in list is n.
Example
Given linked list: 1->2->3->4->5->null, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5->null.
Challenge
O(n) time
~~~
### 题解
简单题，使用快慢指针解决此题，需要注意最后删除的是否为头节点。让快指针先走`n`步，直至快指针走到终点，找到需要删除节点之前的一个节点，改变`node->next`域即可。
### 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.
* @param n: An integer.
* @return: The head of linked list.
*/
ListNode *removeNthFromEnd(ListNode *head, int n) {
if (NULL == head || n < 0) {
return NULL;
}
ListNode *preN = head;
ListNode *tail = head;
// slow fast pointer
int index = 0;
while (index < n) {
if (NULL == tail) {
return NULL;
}
tail = tail->next;
++index;
}
if (NULL == tail) {
return head->next;
}
while (tail->next) {
tail = tail->next;
preN = preN->next;
}
preN->next = preN->next->next;
return head;
}
};
~~~
以上代码单独判断了是否需要删除头节点的情况，在遇到头节点不确定的情况下，引入`dummy`节点将会使代码更加优雅，改进的代码如下。
### C++ dummy node
~~~
/**
* 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.
* @param n: An integer.
* @return: The head of linked list.
*/
ListNode *removeNthFromEnd(ListNode *head, int n) {
if (NULL == head || n < 1) {
return NULL;
}
ListNode *dummy = new ListNode(0);
dummy->next = head;
ListNode *preDel = dummy;
for (int i = 0; i != n; ++i) {
if (NULL == head) {
return NULL;
}
head = head->next;
}
while (head) {
head = head->next;
preDel = preDel->next;
}
preDel->next = preDel->next->next;
return dummy->next;
}
};
~~~
### 源码分析
引入`dummy`节点后画个图分析下就能确定`head`和`preDel`的转移关系了。
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