letecode [237] - 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.

题目大意：

　　给定链表中要被删除的节点，将它从链表中删除。

理  解：

　　没有给定链表。参考他人的方法，将该节点后面的一个节点元素值赋值给该节点，再删除该节点。

代 码 C++：

/**
 * 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;
        node->next = node->next->next;
    }
};

运行结果：

　　执行用时 :24 ms, 在所有C++提交中击败了57.73%的用户

　　内存消耗 :9.3 MB, 在所有C++提交中击败了5.97%的用户

标签：Node,node,ListNode,val,list,List,next,letecode,linked
来源： https://www.cnblogs.com/lpomeloz/p/11024049.html