237. Delete Node in a Linked List

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 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the linked list should become 1 -> 2 -> 4 after 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:

唔，这道题我也刷了三遍...

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;}
}

转载于:https://www.cnblogs.com/yrbbest/p/5003811.html

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