237. Delete Node in a Linked List
2015-11-29 00:01
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题目:
Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.
Supposed the linked list is
链接: http://leetcode.com/problems/delete-node-in-a-linked-list/
题解:
没啥可说的,就是干!
Time Complexity - O(1), Space Complexity - in place
二刷:
就是改变当前node的val和next节点,都变成下一个节点的值和reference就可以了。
Java:
三刷:
因为不是delete tail,意味着当前节点非空,并且下一个节点非空。 所以我们可以略去一些边界条件的判断。
Java:
Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.
Supposed the linked list is
1 -> 2 -> 3 -> 4and you are given the third node with value
3, the linked list should become
1 -> 2 -> 4after calling your function.
链接: http://leetcode.com/problems/delete-node-in-a-linked-list/
题解:
没啥可说的,就是干!
Time Complexity - O(1), Space Complexity - in place
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public void deleteNode(ListNode node) { node.val = node.next.val; node.next = node.next.next; } }
二刷:
就是改变当前node的val和next节点,都变成下一个节点的值和reference就可以了。
Java:
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public void deleteNode(ListNode node) { if (node == null || node.next == null) { return; } node.val = node.next.val; node.next = node.next.next; } }
三刷:
因为不是delete tail,意味着当前节点非空,并且下一个节点非空。 所以我们可以略去一些边界条件的判断。
Java:
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public void deleteNode(ListNode node) { node.val = node.next.val; node.next = node.next.next; } }
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