p19

LeetCode p19 Remove Nth Node From End of List 题解

1.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.

题意：

移除链表中倒数第K个数

2.解题思路:

快慢指针定位要移除的位置。加上一个链表头防止空指针情况。

3.代码

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[title] [] [url] [link text]

 
/**
 * 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 start = new ListNode(0);
		start.next = head;

		ListNode fast = start;
		ListNode slow = start;
		for (int i = 0; i < n + 1; i++) {
			fast = fast.next;
		}
		while (fast != null) {
			fast = fast.next;
			slow = slow.next;
		}
		slow.next = slow.next.next;

		return start.next;
    }
}

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4.一些总结：