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.
思路:有了note中的限定之后,解决方案很简单,直接用待删除节点node的下一个节点覆盖node节点就可以了。
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 void deleteNode(ListNode* node) { 12 ListNode *next = node->next; 13 node->val = next->val; 14 node->next = next->next; 15 } 16 };