【LeetCode】 237. Delete Node in a Linked List

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:

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.

寫一個函式能刪除單向串列鏈結中的一個節點(不會是尾巴)，只會給予該節點的存取權。

給予串列鏈結的頭 = [4,5,1,9]，它看起來就像這樣：

注意：

• 該串列鏈結至少擁有兩個元素。
• 所有的節點都不重複。
• 給予的節點不會是尾巴，且它總是串列鏈結中的合法節點。
• 不要回傳任何東西。

Example:

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.

Solution

• 基本的Linked-List練習題。
• 刪除節點x時，通常我們會將x - 1的next指向x + 1，並將節點x刪除。
  • 但因為這題只給予目標節點的存取權，拿不到所謂的x - 1，因為我們換一個做法。
• 我們先將x的值設為x + 1的值，然後直接將x的next指向x + 2，最後將x + 1刪除。
  • 概念是完全一樣的，只是把整體往後移了一步。

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) {
        node->val = node->next->val;
        auto temp = node->next;
        node->next = node->next->next;
        delete(temp);
    }
};

tags: LeetCode C++