leetCode(9):Remove Nth Node From End of List

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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) {
        if(!head)
    		return NULL;
    	ListNode* fast=head;
    	ListNode* slow=head;
    	int i=0;
    	while(i<n+1 && fast)
    	{
    		i++;
    		fast=fast->next;
    	}
    	if(!fast && i==n)
    	{
    		ListNode* toBeDeleted=head;
    		ListNode* newHead=toBeDeleted->next;
    		delete toBeDeleted;
    		toBeDeleted=NULL;
    		return newHead;
    	}
    	if(!fast && i<n+1)
    		return head;
    	while(fast)
    	{
    		slow=slow->next;
    		fast=fast->next;
    	}
    	ListNode* toBeDeleted=slow->next;
    	slow->next=toBeDeleted->next;
    	delete toBeDeleted;
    	toBeDeleted=NULL;
    	return head;	
    }
};

leetCode(9):Remove Nth Node From End of List

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原文地址：http://blog.csdn.net/walker19900515/article/details/46543773