Algorithm
LeetCode算法题
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 Explanation: 342 + 465 = 807.
Java 程序实现
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if (l1 == null && l2 == null) {
throw new IllegalArgumentException("illegal argument");
}
if (l1 == null) {
return l2;
}
if (l2 == null) {
return l1;
}
// head
int low = 0;
int t = l1.val + l2.val;
if (t > 9) {
low = 1;
t -= 10;
}
ListNode head = new ListNode(t);
ListNode first = l1.next;
ListNode second = l2.next;
ListNode n = head;
while (first != null || second != null || low == 1) {
int x = first == null ? 0 : first.val;
int y = second == null ? 0 : second.val;
int total = x + y + low;
// reset
low = 0;
if (total <= 9) {
n.next = new ListNode(total);
} else {
n.next = new ListNode(total - 10);
low = 1;
}
// forward
first = first == null ? null : first.next;
second = second == null ? null : second.next;
n = n.next;
}
return head;
}
}
Review
详细介绍了分布式锁使用场景，Redis提供的Redlock分布式锁实现的鸡肋之处，代价大，要求高（时延），结论是一般场景用单节点的Redis锁，特殊场景基于zookeeper使用分布式锁。
Tip
错误越早发现，代价越小，如果从模型设计上就有问题，后续的实现基本就是补丁套补丁，丑陋不堪，费时耗力效果差，作为架构师，其设计能力至关重要。
