Question

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• lintcode: (174) Remove Nth Node From End of List

Given a linked list, remove the nth node from the end of list and return its head.

Note
The minimum number of nodes in list is n.

Example
Given linked list: 1->2->3->4->5->null, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5->null.

Challenge
O(n) time

题解

简单题， 使用快慢指针解决此题，需要注意最后删除的是否为头节点。让快指针先走 n 步，直至快指针走到终点，找到需要删除节点之前的一个节点，改变 node->next 域即可。

C++

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *   int val;
 *   ListNode *next;
 *   ListNode(int val) {
 *     this->val = val;
 *     this->next = NULL;
 *   }
 * }
 */
class Solution {
public:
  /**
   * @param head: The first node of linked list.
   * @param n: An integer.
   * @return: The head of linked list.
   */
  ListNode *removeNthFromEnd(ListNode *head, int n) {
    if (NULL == head || n < 0) {
      return NULL;
    }

    ListNode *preN = head;
    ListNode *tail = head;
    // slow fast pointer
    int index = 0;
    while (index < n) {
      if (NULL == tail) {
        return NULL;
      }
      tail = tail->next;
      ++index;
    }

    if (NULL == tail) {
      return head->next;
    }

    while (tail->next) {
      tail = tail->next;
      preN = preN->next;
    }
    preN->next = preN->next->next;

    return head;
  }
};

以上代码单独判断了是否需要删除头节点的情况，在遇到头节点不确定的情况下，引入 dummy 节点将会使代码更加优雅，改进的代码如下。

C++ dummy node

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *   int val;
 *   ListNode *next;
 *   ListNode(int val) {
 *     this->val = val;
 *     this->next = NULL;
 *   }
 * }
 */
class Solution {
public:
  /**
   * @param head: The first node of linked list.
   * @param n: An integer.
   * @return: The head of linked list.
   */
  ListNode *removeNthFromEnd(ListNode *head, int n) {
    if (NULL == head || n < 1) {
      return head;
    }

    ListNode dummy(0);
    dummy.next = head;
    ListNode *preDel = dummy;

    for (int i = 0; i != n; ++i) {
      if (NULL == head) {
        return NULL;
      }
      head = head->next;
    }

    while (head) {
      head = head->next;
      preDel = preDel->next;
    }
    preDel->next = preDel->next->next;

    return dummy.next;
  }
};

源码分析

引入 dummy 节点后画个图分析下就能确定 head 和 preDel 的转移关系了。