LeetCode解题报告--Remove Nth Node From End of List

**题目：**Remove Nth Node From End of List
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.
原题链接：https://leetcode.com/problems/remove-nth-node-from-end-of-list/
分析：改题要求从List删除倒数第N个元素。
解题思路找到倒数第N个元素，修改指针的指向，那么问题来了，如何找到第N个元素？
1）定义两个）ListNode对象（C/C++ 指针：p,q,开始指向List头指针head,
2）p指针开始移动N个位置，此时p-q的长度差为N，现在很自然的想法，整体移动p,q两个对象，知道p移动到最后一个原素（尾指针），此时，q指向的下一个元素即为要删除的（倒数第N个）元素。

Java Accepted 代码：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode p = head;//Faster pointer
        ListNode q = head;//Slower pointer

        if(head.next == null){
            return null;
        }

        //Faster pointer first travel n nodes
        for(int i = 0;i < n;i ++){
            p = p.next;
        }
        if(p == null){
            head = q.next;
            return head;
        }
        //When p.next == null, q points the node that required
        while(p.next != null){
            p = p.next;
            q = q.next;
        }

        q.next = q.next.next;
        return head;
    }
}

相关代码放在个人github:https://github.com/gannyee/LeetCode/