LeetCode---Remove Nth Node From End of List

标签：leetcode   双指针   链表   

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(n<1 || head == NULL)
            return head;
        ListNode* pre = head;
        int i=0;
        while(i<n && pre != NULL)
        {
            pre = pre->next;
            i++;
        }
        if(i<n)
            return head;
        if(pre == NULL)
            return head->next;
        ListNode* back = head;
        ListNode* save;
        while(pre != NULL)
        {
            pre = pre->next;
            save = back;
            back = back->next;
        }
        save->next = back->next;
        delete back;
        return head;
    }
};

LeetCode---Remove Nth Node From End of List

标签：leetcode   双指针   链表   

原文地址：http://blog.csdn.net/shaya118/article/details/42537241