My code:
/**
* 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)
return;
ListNode temp = node;
while (temp.next != null) {
temp.val = temp.next.val;
if (temp.next.next == null) {
temp.val = temp.next.val;
temp.next = null;
break;
}
else
temp = temp.next;
}
}
}
这次作业不难，一开始想着怎么可能删除结点呢，后来觉得是改数值，这样就简单很多了。
**
总结：貌似是第50题。
只不过最近做了好多 easy 题目啊。。。
**
Anyway, Good luck, Richardo!
My code:
/**
* 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)
return;
ListNode pre = node;
while (node.next != null) {
pre = node;
ListNode next = node.next;
node.val = next.val;
node = next;
}
pre.next = null;
return;
}
}
题目不难。
现在才知道，原来还有一种效率更高的做法。比如，
1 2 3 4 5 6
我要删除 3
那么将 3 改成 4 然后让该结点指向5
1 2 4 5 6
那么复杂度就是 O(1)了。
妙！
Anyway, Good luck, Richardo!
My code:
/**
* 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) {
return;
}
ListNode dummy = new ListNode(-1);
dummy.next = node;
ListNode curr = dummy.next;
ListNode pre = dummy;
while (curr.next != null) {
curr.val = curr.next.val;
curr = curr.next;
pre = pre.next;
}
pre.next = null;
}
}
突然发现这个做法好蠢。看了以前的总结才记起还有一种 O(1)的做法。
/**
* 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) {
return;
}
node.val = node.next.val;
node.next = node.next.next;
}
}
Anyway, Good luck, Richardo! -- 08/15/2016
