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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.
/**
* 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 size = 0;
if (n <= 0 || !head)
return head;
ListNode *list1 = head, *list2 = head;
int i = 0;
for (i = 0; i < n; i++)
{
if (!list2)
break;
list2 = list2->next;
}
if (i != n) // n大于链表长度
return head;
if (!list2)
{
ListNode* temp = head;
head = head->next;
delete temp;
return head;
}
while (list2->next)
{
list1 = list1->next;
list2 = list2->next;
}
ListNode* temp = list1->next;
list1->next = temp->next;
delete temp;
return head;
}
};
19.Remove Nth Node From End of List
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原文地址：http://my.oschina.net/hejunsen/blog/421763
