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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.
題意:
刪除鏈表指定節點。前提是只傳刪除節點給函數。但不包括刪除尾節點。
思路:
由于沒有鏈表前節點的存在,所以刪除鏈表時無法改變前節點的指向。但是鏈表值是int型的,所有可以把當前節點的val和下個節點的val交換。然后刪除下個節點即可。
/** * 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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