【剑指 Offer 18. 删除链表的节点】

 【解题思路】

          找到待输出的节点，将前一个节点的指针，指向它后面的节点。

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