https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/
题目
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.
题目翻译
题目解析
参考答案
/**
* 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) {
int iLen = 0;
ListNode * p = head;
while(p != NULL)
{
iLen++;
p = p->next;
}
p = head;
for (int i=0; i<iLen-n-1; i++)
{
p = p->next;
}
if((p == head) && (n == iLen))
{
head = p->next;
delete p;
return head;
}
ListNode * pstDel = p->next;
if(pstDel->next == NULL)
{
p->next = NULL;
}
else
{
p->next = pstDel->next;
}
delete pstDel;
return head;
}
};
这个方法遍历的2遍。
如果要遍历1遍，则需要拿一个指针p1指向head，p1先向前走n步， 然后再拿一个指针p2指向head，p1和p2同时向前，p1走到尾部时，p2就指向了要删除的节点。
