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删除指定节点Remove Nth Node From End of List

时间:2015-01-23 19:48:54      阅读:241      评论:0      收藏:0      [点我收藏+]

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.

 

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 Linked List Two Pointers
 
/**
 * 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 *temp = head;
        int len=0,m;
        while(temp){
            ++len;
            temp=temp->next;
        }
        if(n==len)
            return head->next;
        m=len-n;
        ListNode *temp1=head;
        while(m>1){
            temp1=temp1->next;
            --m;
        }
        temp1->next=temp1->next->next;
        return head;
            
    }
};

 

删除指定节点Remove Nth Node From End of List

原文:http://www.cnblogs.com/li303491/p/4244796.html

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