237. Delete Node in a Linked List

Iterate

class Solution {
    public void deleteNode(ListNode node) {
        ListNode pre = null;
        ListNode curr = node;
        
        while (curr.next != null) {
            curr.val = curr.next.val;
            pre = curr;
            curr = curr.next;
        }
        
        // now: curr.next is null, curr is tail
        pre.next = null;
    }
}

Iterate

class Solution {
    public void deleteNode(ListNode node) {
        node.val = node.next.val;
        node.next = node.next.next;
    }
}

一般的删除节点操作需要通过修改前驱节点的 next 指针来实现，而本题只提供了待删除节点，所以采用了复制 next 节点值至当前节点并删除 next 节点的方式来实现。开始我写出的解法为将 node 节点后的节点值依次覆盖至前驱节点然后再移除尾节点，但是没有该解法简洁。

Reference

237. Delete Node in a Linked List