237 - Delete Node in a Linked List

最新推荐文章于 2019-07-18 09:57:21 发布

原创最新推荐文章于 2019-07-18 09:57:21 发布 · 176 阅读

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本文介绍了一个简单的算法，用于在单链表中删除指定节点（除尾节点外），仅需访问该节点。通过修改节点值及指向下一个节点的指针实现删除操作。

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Problem

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.

Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void deleteNode(ListNode* node) {
        if (node == NULL) {
            return;
        }
        node->val = node->next->val;
        node->next = node->next->next;
    }
};