【一天一道LeetCode】#19. Remove Nth Node From End of List
一天一道LeetCode系列
(一)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.
(二)解题
剑指上面的经典题目了。两个指针,一个指向head,一个指向正数第K-1个,同时移动,知道第K-1的指针指向NULL。则指向head的数就是倒数第K个。
删除节点的时候要区分头节点和其他节点。
/**
* 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(n == 0) return head;
ListNode* ptemp = head;
for(int i = 0 ; i < n ;i++)
{
ptemp = ptemp->next;
}
ListNode* pNth = head;
ListNode* pre = head;
while(ptemp!= NULL)
{
pre = pNth;
ptemp = ptemp->next;
pNth = pNth->next;
}
if(pNth == head)//如果是头节点
{
head = head->next;
}
else pre->next = pNth->next;//非头节点
free(pNth);
return head;
}
};
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