LeetCode题解：Remove Nth Node from End of List

Remove Nth Node From End of List

Given a linked list, remove the n^th 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.

思路：

用两个指针。第一个指针先前进n个结点，然后第二个指针跟进。第一个指针走到队列的尾端的时候，第二个指针刚好落到要删除结点的前一个位置。要考虑删除的结点刚好是第一个结点的情况。

题解：

/** * 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* fwd_iter = head;                for(int i = 0; i < n; ++i)            fwd_iter = fwd_iter->next;                ListNode* del_iter = head;                if (fwd_iter == nullptr)    // deleting first ptr        {            head = head->next;            delete del_iter;            return head;        }                while(fwd_iter->next != nullptr)        {            fwd_iter = fwd_iter->next;            del_iter = del_iter->next;        }                fwd_iter = del_iter->next;        del_iter->next = fwd_iter->next;        delete fwd_iter;        return head;    }};