Leetcode 19. Remove Nth Node From End of List

Given a linked list, remove the nth node from the end of list and return its head.

For example,

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Try to do this in one pass.

题目大意：删除链表中倒数第n个节点。

题目分析：使用双指针相聚n-1，当第二个指针遍历到链表尾部则第一个指针就是要删除的节点指针。这里有个问题，因为要删除节点，所以不添加头结点的话，要把删除第一个节点，和其他节点区分开来（第一个节点没有前驱）。如果添加则可以统一处理。下面分别是两种代码，运行时间都是4ms

方法一（不添加头节点）：

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
if(head==NULL||head->next==NULL) return NULL;
ListNode* first=head;
ListNode* second=head;
for(int i=0;i<n-1;i++)
second=second->next;
if(second->next==NULL){//删除第一个节点
head=head->next;
return head;
}
while(second->next->next){
second=second->next;
first=first->next;
}
first->next=first->next->next;
return head;
}
};方法二（添加头节点）：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
if(head==NULL) return NULL;
ListNode h(INT_MAX);
h.next=head;

ListNode *pfirst=&h;
ListNode *psecond=&h;
ListNode *temp;//释放被删除的空间

for(int i=0;i<n;i++)
psecond=psecond->next;
while(psecond->next){
psecond=psecond->next;
pfirst=pfirst->next;
}
temp=pfirst->next;
pfirst->next=pfirst->next->next;
delete temp;
return h.next;
}
};