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.
分析
题目给了一个链表的一维数组,和一个整数n,要求删除从末尾开始数第n个节点,然后返回链表的头。做法是,用两个指针,一个fast指针,一个solw指针,让slow指针永远慢fast指针n个节点。fast走完后,再用slow指针来删除需要删除的节点
代码
/**
* 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) {
ListNode* h = new ListNode(0);
h->next = head;
ListNode* slow = h;
ListNode* fast = h;
while(fast->next!=NULL){
if(n<=0){
slow = slow->next;
}
fast= fast->next;
n--;
}
if(slow->next!=NULL){
slow->next = slow->next->next;
}
return h->next;
}
};
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