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.
由于删除头结点时要改变头结点的指针，所以传入的是引用或者使用二级指针。提示中给的只是传入一级指针，要是删除头结点，始终不能删掉。（注意要改变一个数据结构时，要传二级指针或者引用，树中的很多操作都是这样的）
C++实现代码如下：
#include<iostream> #include<new> using namespace std; //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 *p=head; ListNode *q=head; ListNode *pre=head; int i=0; while(q&&i<n) { q=q->next; i++; } //说明没有n个元素 if(i<n) return head; while(q) { pre=p; p=p->next; q=q->next; } if(p==head) { head=head->next; delete p; } else { pre->next=p->next; delete p; } return head; } void createList(ListNode *&head) { ListNode *p=NULL; int i=0; int arr[10]= {10,9,8,7,6,5,4,3,2,1}; for(i=0; i<5; i++) { if(head==NULL) { head=new ListNode(arr[i]); if(head==NULL) return; } else { p=new ListNode(arr[i]); p->next=head; head=p; } } } }; int main() { Solution s; ListNode *L=NULL; s.createList(L); ListNode *head=L; while(head) { cout<<head->val<<" "; head=head->next; } cout<<endl; s.removeNthFromEnd(L,5); while(L) { cout<<L->val<<" "; L=L->next; } }
运行结果：
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