237. Delete Node in a Linked List&在O(1)时间删除链表结点
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题目
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.
解答
在这个问题中,无法访问链表的头结点,只能访问待删除的结点。方法很简单,直接将后继结点的数据复制到待删除的结点,然后删掉待删除结点的后继结点。
/*struct ListNode { int val; struct ListNode *next; ListNode(int x) : val(x), next(NULL) {}};*/class Remove {public: bool removeNode(ListNode* pNode) { // write code here if(!pNode || !pNode->next) { return false; } ListNode* next = pNode->next; pNode->val = next->val; pNode->next = next->next; delete next; return true; }};
如果待删除的结点是链表的尾结点,那么该问题无解
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