LeetCode-Remove Nth Node From End of List

标签：leetcode   remove nth node from   

题目链接：https://leetcode.com/problems/remove-nth-node-from-end-of-list/

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.

解题思路：要求删除倒数第N个数，由于链表的顺序搜索特性。所以改变思想，引用两个链表 L1，L2。L1先顺序搜索N个数，然后L1，L2再一起搜索直到L1到底，删除L2的这个点就行。

class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        ListNode *point = head;
        ListNode *answers = head;
        ListNode *change = head;
        for (int i = 0; i < n; i++) {
            if (point->next == NULL) return answers->next;
            point = point->next;
        }
        while(point->next != NULL) {
            head = head->next;
            point = point->next;
        }
        head->next = head->next->next;
        return answers;
    }
};

LeetCode-Remove Nth Node From End of List

标签：leetcode   remove nth node from   

原文地址：http://blog.csdn.net/vanish_dust/article/details/44150571