LeetCode 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.

Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the linked list should become 1 -> 2 -> 4 after calling your function.

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这道题很简单，给定一个节点，要求删除当前节点。但是你可能会问，前一个节点指针都没有，怎么删除当前节点啊。这里可以抖机灵，把当前节点赋值为下一个节点，然后下一个节点就被抛弃了。等价于把当前节点删除了：
删除节点2之前：

将节点2赋值为节点3：

这就等于把原来的节点2删掉了。

AC CODE：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */void deleteNode(struct ListNode* node) {    if(node->next == NULL) return; //注意最后一个节点是删不掉的    node->val = node->next->val;    node->next = node->next->next;}