leetcode Remove Nth Node From End of List

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本文介绍了一种高效算法，用于从链表中删除倒数第N个节点，并确保只遍历一次链表。该算法使用两个指针，先让第一个指针向前移动N步，然后同时移动两个指针，直到第一个指针到达链表末尾。此时，第二个指针指向待删除节点的前一个节点，进而实现删除操作。

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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.

Use two pointers and the first pointer move forwards n steps first.

/**
 * 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) {
    // IMPORTANT: Please reset any member data you declared, as
    // the same Solution instance will be reused for each test case.
    ListNode *dummy = new ListNode(0), *first = dummy, *second = dummy, *last = NULL, *res = NULL;
    dummy->next = head;
    for (int i = 0; i < n; ++i)
      first = first->next;
    while (first != NULL) {
      first = first->next;
      last = second;
      second = second->next;
    }
    last->next = second->next;
    delete second;
    res = dummy->next;
    delete dummy;
    return res;
  }
};