LeetCode203.Remove Linked List Elements（ 删除链表中的节点）

Remove all elements from a linked list of integers that have value val. 
　　Example 
　　Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6 
　　Return: 1 --> 2 --> 3 --> 4 --> 5 

删除链表中等于给定值 val 的所有节点

示例:输入: 1->2->6->3->4->5->6, val = 6 ；输出: 1->2->3->4->5

方法一：使用虚拟头节点

public class Solution {
	public ListNode removeElements(ListNode head, int val) {
		ListNode dummyHead = new ListNode(-1);
		dummyHead.next = head;
		ListNode prev = dummyHead;
		while (prev.next != null) {
			if (prev.next.val == val)
				prev.next = prev.next.next;
			else
				prev = prev.next;
		}
		return dummyHead.next;
	}
}

方法二：不使用虚拟头节点

public class Solution {
	public ListNode removeElements(ListNode head, int val) {
		while (head != null && head.val == val)
			head = head.next;
		if (head == null)
			return head;
		ListNode prev = head;
		while (prev.next != null) {
			if (prev.next.val == val)
				prev.next = prev.next.next;
			else
				prev = prev.next;
		}
		return head;
	}
}

方法三：使用递归

public class Solution {
	public ListNode removeElements(ListNode head, int val) {
		if (head == null)
			return null;
		ListNode res = removeElements(head.next, val);
		if (head.val == val)
			return res;
		else {
			head.next = res;
			return head;
		}
	}
}

测试代码

public class ListNode {
	public int val;
	public ListNode next;

	public ListNode(int x) {
		val = x;
	}
	// 链表节点的构造函数
	// 使用arr为参数，创建一个链表，当前的ListNode为链表头结点
	public ListNode(int[] arr) {

		if (arr == null || arr.length == 0)
			throw new IllegalArgumentException("arr can not be empty");

		this.val = arr[0];
		ListNode cur = this;
		for (int i = 1; i < arr.length; i++) {
			cur.next = new ListNode(arr[i]);
			cur = cur.next;
		}
	}
	// 以当前节点为头结点的链表信息字符串
	@Override
	public String toString() {

		StringBuilder s = new StringBuilder();
		ListNode cur = this;
		while (cur != null) {
			s.append(cur.val + "->");
			cur = cur.next;
		}
		s.append("NULL");
		return s.toString();
	}
}

public class Test {
	public static void main(String[] args) {
		int[] nums = { 1, 2, 6, 3, 4, 5, 6 };
		ListNode head = new ListNode(nums);
		System.out.println(head);
		ListNode res = (new Solution()).removeElements(head, 6);
		System.out.println(res);
	}
}

运行结果