Remove Nth Node From End of List ——解题报告

标签：leetcode   链表   两指针法   删除节点   

    【题目】

    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.

    【分析】

    两指针法。先让first领先n-1步，然后first和second一起移动，当first到底的时候，second即为倒数第k个节点。题目需要删除，所以我们再额外记录second的前节点。

    注意其中特殊情况的考虑。

    【代码】

/**
 * 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) {
        if(head == NULL || n <= 0)
            return head;
        
        ListNode* first = head;
        ListNode* second = head;
        ListNode* pre = NULL;
        for(int i = 0; i < n - 1; i++) // 中间间隔N-1的距离
            first = first->next;
        while(first->next != NULL)
        {
            pre = second;  // pre是倒数第k个节点的前一个节点，用来删除
            second = second->next;  // second是倒数第k个节点
            first = first->next;
        }
        
        if(pre == NULL)  // 如果倒数第k个节点就是head，那么pre会保留原来的NULL，这时head直接是倒数第k个节点second的下一个节点
            head = second->next;
        else
            pre->next = second->next;
        
        return head;
    }
};

Remove Nth Node From End of List ——解题报告

标签：leetcode   链表   两指针法   删除节点   

原文地址：http://blog.csdn.net/puqutogether/article/details/45575703