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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.
题意:
删除链表指定节点。前提是只传删除节点给函数。但不包括删除尾节点。
思路:
由于没有链表前节点的存在,所以删除链表时无法改变前节点的指向。但是链表值是int型的,所有可以把当前节点的val和下个节点的val交换。然后删除下个节点即可。
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ void deleteNode(struct ListNode* node) { if ( node->next == NULL ) { return; } int tmp = 0; tmp = node->val; node->val = node->next->val; node->next->val = tmp; struct ListNode *list = node->next; node->next = node->next->next; free(list); }
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