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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 <= 500 <= 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));}
}
