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203.移除链表元素

题目描述

原题

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:

  • 列表中的节点在范围 [0, 104]
  • 1 <= Node.val <= 50
  • 0 <= k <= 50

题解

递归

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

迭代

/**
 * 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) {
        //创建一个模拟的头节点
        ListNode dummyNode = new ListNode();
        dummyNode.next = head;
        head = dummyNode;
        while(head!=null){
            ListNode next = head.next;
            //循环找到值不相等的next节点
            while(next!=null&&next.val == val){
                next = next.next;
            }
            //指向next
            head.next = next;
            head = next;
        }
        return dummyNode.next;
    }
}
//官方代码优化
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        ListNode dummyHead = new ListNode(0);
        dummyHead.next = head;
        ListNode temp = dummyHead;
        while (temp.next != null) {
            if (temp.next.val == val) {
                temp.next = temp.next.next;
            } else {
                temp = temp.next;
            }
        }
        return dummyHead.next;
    }
}