数据结构与算法——Java实现 8.习题——移除链表元素（值）

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                                                                        —— 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;}@Overridepublic 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));}
}