祝福你有前路坦途的好运，更祝愿你能保持内心光亮

纵有风雨，依然选择勇敢前行

                                                                        —— 24.9.22

203. 移除链表元素

给你一个链表的头节点 head 和一个整数 val ，请你删除链表中所有满足 Node.val == val 的节点，并返回 新的头节点 。

示例 1：

输入：head = [1,2,6,3,4,5,6], val = 6
输出：[1,2,3,4,5]

示例 2：

输入：head = [], val = 1
输出：[]

示例 3：

输入：head = [7,7,7,7], val = 7
输出：[]

提示：

• 列表中的节点数目在范围 [0, 104] 内
• 1 <= Node.val <= 50
• 0 <= val <= 50

方法1

思路

定义哨兵节点，定义两个指针，指针1指向头结点，指针2指向头结点的下一个结点，进行循环，比较指针2指向的结点是否等于要删除的结点，如果等于，则接着后移进行遍历，直至指针2指向空，遍历结束

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public static ListNode removeElements(ListNode head, int val) {
        ListNode s = new ListNode(-1,head);
        ListNode p1 = s;
        ListNode p2 = s.next;
        while (p2 != null) {
            if (p2.val == val) {
                p1.next = p2.next;
                p2 = p2.next;

            }else {
                p1 = p2;
                p2 = p2.next;
            }
        }
        return s.next;
    }
}

方法2

思路

递归函数负责返回:从当前节点开始，完成删除的的链表

        1.若当前节点与目标相等，应该返回下一个节点递归结果

        2.若当前节点与目标不等，应该返回当前节点，但当前节点的 next 应该更新

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        if (head == null) {
            return head;
        }
        head.next = removeElements(head.next, val);
        return head.val == val ? head.next : head;
    }
}

完整代码

ListNode类定义

package Day9ListPractice;

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

    public ListNode(int val, ListNode next) {
        this.val = val;
        this.next = next;
    }

    // 可变长参数
    public static ListNode of(int...numbers) {
        ListNode head = new ListNode(0, null);
        ListNode current = head;
        for (int number : numbers) {
            current.next = new ListNode(number, null);
            current = current.next;
        }
        return head;
    }


    @Override
    public String toString() {
        StringBuilder sb = new StringBuilder(64);
        sb.append("[");
        ListNode p = this;
        while (p != null) {
            sb.append(p.val);
            if (p.next != null) {
                sb.append(",");
            }
            p = p.next;
        }
        sb.append("]");
        return sb.toString();
    }
}

方法函数 

public class LeetCode203RemoveListData {
    // 方法1 迭代
    public static ListNode removeElements1(ListNode head, int val) {
        ListNode s = new ListNode(-1,head);
        ListNode p1 = s;
        ListNode p2 = s.next;
        while (p2 != null) {
            if (p2.val == val) {
                p1.next = p2.next;
                p2 = p2.next;

            }else {
                p1 = p2;
                p2 = p2.next;
            }
        }
        return s.next;
    }

    // 方法2 递归
    public ListNode removeElements2(ListNode head, int val) {
        if (head == null) {
            return head;
        }
        head.next = removeElements2(head.next, val);
        return head.val == val ? head.next : head;
    }

    public static void main(String[] args) {
        ListNode head = ListNode.of(1,2,3,4,5,6,7,8);
        System.out.println(head);
        System.out.println(new LeetCode203RemoveListData().removeElements1(head, 1));
        System.out.println(new LeetCode203RemoveListData().removeElements2(head, 7));
    }
}