LeetCode:Remove Nth Node From End of List

题目链接

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.

主要难点是通过一趟遍历寻找链表倒数第k个元素，具体见代码注释                            本文地址

/**
 * 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) {
        //快指针先走n步，然后快慢指针一起走，块指针指向尾节点时，慢指针指向倒数第n个节点
        ListNode* fast = head, *slow = head, *slowpre = NULL;
        for(int i = 1; i < n; i++)fast = fast->next;
        while(fast->next)
        {
            fast = fast->next;
            slowpre = slow;
            slow = slow->next;
        }
        if(slow == head)
            head = head->next;
        else
            slowpre->next = slow->next;
        delete slow;
        return head;
    }
};

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