237. Delete Node in a Linked List

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:

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.237. Delete Node in a Linked List

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删掉单向链表中的某个指定的节点，这个节点不是尾节点。 本科面试的时候遇到过这个题，当时好像没做出来…因为要删掉某个节点，肯定要知道其前驱节点prior，然后prior->next=prior->next->next。但是这一题只让访问当前节点，单向链表又无法获取其前驱节点，所以当时想了很久都没解出来。 今天灵光一闪，只能访问当前节点，又得不到其前驱节点，唯一可以知道的是其后继节点，所以我们可以不用真正的删掉当前节点，可以把后继节点的值赋给当前节点，然后删掉后继节点！ 想到这一点就很简单了，也许这也是为什么题目说不包括尾节点，因为删掉尾节点不能用上述方法，只能用其前驱节点删除。 完整代码如下：

class Solution {
public:
    void deleteNode(ListNode* node)
    {
        node->val = node->next->val;
        ListNode *no_use=node->next;
        node->next = node->next->next;
        delete no_use;
    }
};

本代码提交AC，用时13MS。