203. 移除链表元素

题目来源：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：[]

 

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} val
 * @return {ListNode}
 */
var removeElements = function(head, val) {
    let cur = head;
    let prev = null;
    while(cur != null){
        if(cur.val === val){
            if(prev === null){
                head = head.next;
            }else{
                prev.next = cur.next;
            }
        }else{
            prev = cur;
        }
        cur = cur.next;
    }
    return head;
};

 Python3 

# 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:
      if head is None:
        return head
      
      head.next  = self.removeElements(head.next, val)

      if head.val == val:
        headNew = head.next
      else:
        headNew = head

      return headNew

class Solution:
    def removeElements(self, head: ListNode, val: int) -> ListNode:
      cur = head
      pre = None
      while cur is not None:
        if cur.val == val:
          if pre is None:
            head = head.next
          else:
            pre.next = cur.next
        else:
          pre = cur
        
        cur = cur.next

      return head

 

提示：

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