【203】移除链表元素

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一、题目

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

二、题解

1、不使用虚拟头节点的方法

public class Solution {

    /**
     * 没有使用虚拟头节点的方法
     *
     * @param head
     * @param val
     * @return
     */
    public ListNode removeElements(ListNode head, int val) {
        // 删除掉从 head 开始都是目标数据的节点
        while (head != null && head.val == val) {
            head = head.next;
        }

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

        // 头部的目标数据删完了，删除头节点后面的为目标数据的节点
        ListNode prev = head;
        while (prev.next != null) {
            if (prev.next.val == val) {
                prev.next = prev.next.next;
            } else {
                prev = prev.next;
            }
        }

        return head;
    }

    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 result = new Solution().removeElements(head, 6);
        System.out.println(result);
    }

    private static class ListNode {

        public int val;
        public ListNode next;

        public ListNode(int x) {
            val = x;
        }

        public ListNode(int[] array) {
            if (array == null || array.length == 0) {
                throw new IllegalArgumentException("array cannot be empty");
            }

            val = array[0];
            ListNode current = this;

            for (int i = 1; i < array.length; i++) {
                current.next = new ListNode(array[i]);
                current = current.next;
            }
        }

        @Override
        public String toString() {
            StringBuilder result = new StringBuilder();
            ListNode current = this;
            while (current != null) {
                result.append(current.val).append("->");
                current = current.next;
            }
            result.append("null");
            return result.toString();
        }
    }
}

运行结果：

1->2->6->3->4->5->6->null
1->2->3->4->5->null

2、使用虚拟头节点的方法

public class Solution {

    /**
     * 使用虚拟头节点的方法
     *
     * @param head
     * @param val
     * @return
     */
    public ListNode removeElements(ListNode head, int val) {
        // 虚拟头节点
        ListNode dummyHead = new ListNode(-1);
        dummyHead.next = head;

        // 因为所有的节点都具有前一个节点，所以可以直接循环
        ListNode prev = dummyHead;
        while (prev.next != null) {
            if (prev.next.val == val) {
                prev.next = prev.next.next;
            } else {
                prev = prev.next;
            }
        }

        // 因为有可能头节点已经被删除了，但是头节点元素还有值，所以这里要返回虚拟头节点的 next
        return dummyHead.next;
    }

    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 result = new Solution().removeElements(head, 6);
        System.out.println(result);
    }

    private static class ListNode {

        public int val;
        public ListNode next;

        public ListNode(int x) {
            val = x;
        }

        public ListNode(int[] array) {
            if (array == null || array.length == 0) {
                throw new IllegalArgumentException("array cannot be empty");
            }

            val = array[0];
            ListNode current = this;

            for (int i = 1; i < array.length; i++) {
                current.next = new ListNode(array[i]);
                current = current.next;
            }
        }

        @Override
        public String toString() {
            StringBuilder result = new StringBuilder();
            ListNode current = this;
            while (current != null) {
                result.append(current.val).append("->");
                current = current.next;
            }
            result.append("null");
            return result.toString();
        }
    }
}

运行结果：

1->2->6->3->4->5->6->null
1->2->3->4->5->null

3、使用递归解决

public class Solution {

    /**
     * 使用递归解决
     *
     * @param head
     * @param val
     * @return
     */
    public ListNode removeElements(ListNode head, int val) {
        if (head == null) {
            return null;
        }

        head.next = removeElements3(head.next, val);
        return head.val == val ? head.next : head;
    }

    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 result = new Solution().removeElements(head, 6);
        System.out.println(result);
    }

    private static class ListNode {

        public int val;
        public ListNode next;

        public ListNode(int x) {
            val = x;
        }

        public ListNode(int[] array) {
            if (array == null || array.length == 0) {
                throw new IllegalArgumentException("array cannot be empty");
            }

            val = array[0];
            ListNode current = this;

            for (int i = 1; i < array.length; i++) {
                current.next = new ListNode(array[i]);
                current = current.next;
            }
        }

        @Override
        public String toString() {
            StringBuilder result = new StringBuilder();
            ListNode current = this;
            while (current != null) {
                result.append(current.val).append("->");
                current = current.next;
            }
            result.append("null");
            return result.toString();
        }
    }
}

运行结果：

1->2->6->3->4->5->6->null
1->2->3->4->5->null