Leetcode ☞237. Delete Node in a Linked List ☆

237. Delete Node in a Linked List

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Question

Total Accepted: 65936 Total Submissions: 150816 Difficulty: Easy

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.

我的AC（4ms，前98%）：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */void deleteNode(struct ListNode* node) {    struct ListNode *p;        p = node->next;    node->val =  p->val;    node->next = p->next;        free(p);}

分析：

认真读题。提供的是所要删除的那个节点！！！没给各个节点的值也没给长度。

所以把该node的所有“属性”都改成“node->next”的属性，这时原node其实就已经没了（有两个node->next）！然后free掉node->next。

链表的题，画图后就很容易理解。

举个例子：

a->b->c->d->e->····->z，要删c。

那我们其实不是真删了c，（实际是删了c->next）。我们把d的val跟d->next都给c，原链表就变成了

a->b->d(本来是C，值改成了d)->e->····->z 和 d->e->····->z。 然后free掉这个枝杈d。