leetcode笔记 203移除链表元素 python Java

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 。

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

输入：head = [], val = 1 输出：[]

输入：head = [7,7,7,7], val = 7 输出：[]

python解法一：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeElements(self, head: ListNode, val: int) -> ListNode:
        #删除头节点
        while head is not None and head.val == val :
            head = head.next

        #删除中间节点
        current = head
        while current is not None and current.next is not None:
            if current.next.val == val:
                current.next = current.next.next
            else :
                current = current.next

        return head

Java解法一：

/**
 * 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) {
        //删除头节点
        while(head != null && head.val == val){
            head = head.next;
        }

        //删除中间节点
        ListNode current;
        current = head;
        while(current != null && current.next != null){
            if (current.next.val == val){
                current.next = current.next.next;
            }
            else{
                current = current.next;
            }
        }
        return head;
    }
}

python解法二：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeElements(self, head: ListNode, val: int) -> ListNode:
        virtual = ListNode(next = head)
        current = virtual
        while current.next != None:
            if current.next.val == val:
                current.next = current.next.next
            else:
                current = current.next

        return virtual.next

Java解法二：

/**
 * 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) {
        ListNode virtual = new ListNode(0, head);
        ListNode current;
        current = virtual;
        while(current.next != null ){
            if(current.next.val == val){
                current.next = current.next.next;
            }
            else{
                current = current.next;
            }
        }
        return virtual.next;
    }
}