剑指18 删除链表节点

剑指18 删除链表节点

• 此题的一个技巧：为链表附增一个表头节点，方便将所有节点进行统一处理

  /**
  * 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) {
        // 为链表附增一个头节点
        ListNode pHead = new ListNode(-1);
        pHead.next = head;
        ListNode p = pHead;
        while (p.next.val != val)
            p = p.next;
        p.next = p.next.next;
        return pHead.next;
    }
  }

  /**
  * Definition for singly-linked list.
  * type ListNode struct {
  *     Val int
  *     Next *ListNode
  * }
  */
  func deleteNode(head *ListNode, val int) *ListNode {
    preHead := new(ListNode)
    preHead.Next = head
    p := preHead
    for p != nil && p.Next.Val != val {
        p = p.Next
    }
    if p != nil {
        p.Next = p.Next.Next
    }
    return preHead.Next
  }