题解 | #删除链表的节点# 无递归解法

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @param val int整型 
     * @return ListNode类
     */
    public ListNode deleteNode (ListNode head, int val) {
        //保存头节点信息
        ListNode cacheHead = new ListNode(-1);
        cacheHead.next = head;
        
            if(head.val == val){
                head = head.next; 
                return head;                        
            }
            else{
                while(null != head.next){
                    if(head.next.val == val){
                        head.next = head.next.next;
                    }
                    head = head.next;    
                }
                return cacheHead.next;
            }   
    }
}