237.Delete Node in a Linked List

Tags: Easy Linked List

Links: https://leetcode.com/problems/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.

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.

---

/**
 * 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) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        while (node -> next -> next) {
            node -> val = node -> next -> val;
            node = node -> next;
        }
        node -> val = node -> next -> val;
        ListNode *tmp = node -> next;
        node -> next = NULL;
        delete tmp; tmp = NULL;
    }
};

这道题和以往不同，没有给出头节点，所以就没法用pre去记录头节点。删除的效果就是将node后面的链表前移一个位置，那么我们可以将node节点的值替换为node的下一个节点的值，这样就相当于有了头节点，所以删除node的下一个节点即可。