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;
    }
}