[LeetCode]Remove Nth Node From End of List

题目描述

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.

给你一个链表，移除它从后往前数的第n位的节点

解题思路

思路1：

    获取链表长度，移除第len-n个节点；

思路2：

   使用双指针，遍历一遍，移除倒数第n个节点；

代码

思路1代码：

public  ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode node = head;
        
        int length = 0;
        while(node!=null){
        	node = node.next;
        	length++;
        }
        ListNode removeNode = head;
        if(n==length){
        	node = head.next;
        	removeNode = null;
        	return node;
        }
        
        node = head;
        for(int i = 0;i < length - (n+1);i++){
        	node = node.next;
        }
        removeNode = node.next;
        node.next = removeNode.next;
        removeNode = null;
        return head;
	}

思路2代码：

public static ListNode removeNthFromEnd(ListNode head, int n) {
		// 链表为空、n==0或者不符合要求
		if (head == null || n <= 0)
			return null;

		ListNode beforeKthNode = null;
		ListNode pHead = head;
		int i = 0;

		//得到正数第k个节点
		while (pHead != null && i < n) {
			pHead = pHead.next;
			i++;
		}

		if (i < n) {
			return head;
		} else if (i == n && pHead == null) {
			// 移除头节点
			return head.next;
		} else {
			// 得到倒数第k+1个节点
			beforeKthNode = head;
			while (pHead.next != null) {
				pHead = pHead.next;
				beforeKthNode = beforeKthNode.next;
			}

			// 删除倒数第k个节点
			beforeKthNode.next = beforeKthNode.next.next;
		}
		return head;
	}