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;
    }
};