237. Delete Node in a Linked List.md

237. Delete Node in a Linked List

Question

Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

Given linked list -- head = [4,5,1,9], which looks like following:

4 -> 5 -> 1 -> 9

Example 1:

Input: head = [4,5,1,9], node = 5
Output: [4,1,9]
Explanation: You are given the second node with value 5, the linked list
             should become 4 -> 1 -> 9 after calling your function.

Example 2:

Input: head = [4,5,1,9], node = 1
Output: [4,5,9]
Explanation: You are given the third node with value 1, the linked list
             should become 4 -> 5 -> 9 after calling your function.

Note:

• The linked list will have at least two elements.
• All of the nodes' values will be unique.
• The given node will not be the tail and it will always be a valid node of the linked list.
• Do not return anything from your function.

Thinking:

• Method 1:
  • 没有给出链表头，我们将下一个值赋值到当前值。
  • 将下一个结点删除。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public void deleteNode(ListNode node) {
        if(node == null) return;
        node.val = node.next.val;
        node.next = node.next.next;
    }
}

二刷

1. Same as what I did in previous time.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public void deleteNode(ListNode node) {
        ListNode cur = node;
        ListNode pre = null;
        while(cur.next != null){
            cur.val = cur.next.val;
            pre = cur;
            cur = cur.next;
        }
        pre.next = null;
    }
}