[LeetCode] 237. Delete Node in a Linked List

标签：put   http   out   after   lis   oid   xpl   call   element   

Description

Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

Given linked list -- head = [4,5,1,9], which looks like following:

Example 1:

Input: head = [4,5,1,9], node = 5
Output: [4,1,9]
Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.

Example 2:

Input: head = [4,5,1,9], node = 1
Output: [4,5,9]
Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.

Note:

• The linked list will have at least two elements.
• All of the nodes‘ values will be unique.
• The given node will not be the tail and it will always be a valid node of the linked list.
• Do not return anything from your function.

Analyse

从单链表中删除一个节点，但只给要删除的节点的指针，不提供指针的头节点

没有指针的头节点就拿不到要删除的节点的前一个结点，平时删除节点的方法就无法使用了

可以使用类似于数组的删除，将该节点后续的元素前移，然后删除尾节点就行了

void deleteNode(ListNode* node) {
    ListNode* tmp = node;
    while(tmp->next)
    {
        tmp->val = tmp->next->val;
        if (tmp->next->next)
        {
            tmp = tmp->next;
        }
        else
        {
            tmp->next = nullptr;
            return;
        }
    }
}

上面的方法很蠢，完全漠视了链表的特性，每个节点无需保存在连续的内存中，所以后续的节点不用都往前移动

先将要删除的节点node的后面一个节点node->next的值复制到要删除的节点node里，然后将node->next指向node->next->next就行了

void deleteNode(ListNode* node) {
    node->val = node->next->val;
    node->next = node->next->next;
}

[LeetCode] 237. Delete Node in a Linked List

标签：put   http   out   after   lis   oid   xpl   call   element   

原文地址：https://www.cnblogs.com/arcsinw/p/11934659.html