[LeetCode] 237. Delete Node in a Linked List 删除链表的节点
2018-03-14 06:55
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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:
4 -> 5 -> 1 -> 9
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.
写一个函数删除单链表中的一个节点,链表至少有2个元素,所有的节点值是唯一的,给的节点不会是尾部并且是合法的节点,不返回任何值。要求in-place。
解法:先把next节点的值赋给当前节点,在把当前节点的next变成当前节点的next.next。
Java:
public class Solution { public void deleteNode(ListNode node) { node.val = node.next.val; node.next = node.next.next; } }
Python:
# Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def deleteNode(self, node): """ :type node: ListNode :rtype: void Do not return anything, modify node in-place instead. """ node.val = node.next.val node.next = node.next.next
Python:
class Solution: # @param {ListNode} node # @return {void} Do not return anything, modify node in-place instead. def deleteNode(self, node): if node and node.next: node_to_delete = node.next node.val = node_to_delete.val node.next = node_to_delete.next del node_to_delete
C++:
void deleteNode(ListNode* node) { *node = *node->next; }
C++:
class Solution { public: void deleteNode(ListNode* node) { node->val = node->next->val; ListNode *tmp = node->next; node->next = tmp->next; delete tmp; } };
JavaScript:
var deleteNode = function(node) { node.val = node.next.val; node.next = node.next.next; };
Ruby:
def delete_node(node) node.val = node.next.val node.next = node.next.next nil end
类似题目:
[LeetCode] 203. Remove Linked List Elements 移除链表元素
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