訂判斷及所需輸入
題內 L1 把 K1、K2、K3 變 A/B,L2 則 A-B。報 L2 卻一律套 L1,符號就錯。只有實體鍵標記,未選定代碼怎樣讀。
| 情況 | 輸入 JSON |
|---|---|
| 原例測試 | {"layout":"L2","keys":["K1","K2","K3"]} |
跑所附比較
程序只執行所附對應,不代表作業系統、掃描韌體或地區驅動程式;清單收件仍要自己的代碼規則。
| 路徑 | 輸出 JSON |
|---|---|
| 錯誤路徑 | {"layout":"L2","code":"A/B"} |
| 所需路徑 | {"layout":"L2","code":"A-B"} |
import assert from 'node:assert/strict';
const run=function run(i,corrected=true){
const layouts={L1:{K1:'A',K2:'/',K3:'B'},L2:{K1:'A',K2:'-',K3:'B'}};
if(!Object.hasOwn(layouts,i.layout)||i.keys.some(k=>!Object.hasOwn(layouts[i.layout],k)))return {held:true};
return {layout:i.layout,code:i.keys.map(k=>layouts[corrected?i.layout:'L1'][k]).join('')};
};
const cases=[{"input":{"layout":"L2","keys":["K1","K2","K3"]},"corrected":false,"expected":{"layout":"L2","code":"A/B"}},{"input":{"layout":"L2","keys":["K1","K2","K3"]},"corrected":true,"expected":{"layout":"L2","code":"A-B"}},{"input":{"layout":"L1","keys":["K1","K2","K3"]},"corrected":true,"expected":{"layout":"L1","code":"A/B"}},{"input":{"layout":null,"keys":["K1"]},"corrected":true,"expected":{"held":true}}];
for(const test of cases){const actual=await run(test.input,test.corrected);assert.deepEqual(actual,test.expected);console.log(actual);}
按邊界依據作判斷
L1 正當結果是 A/B;配置未知或鍵未有對應便停。
- 核所選配置
- 對整段鍵序列
- 核最後代碼
不能用假設對應推真正設備配置。
| 條件 | 輸入 JSON | 預期輸出 JSON |
|---|---|---|
| 沿第一套配置 | {"layout":"L1","keys":["K1","K2","K3"]} | {"layout":"L1","code":"A/B"} |
| 配置欠缺 | {"layout":null,"keys":["K1"]} | {"held":true} |
原例 L2 應得哪個代碼?
查看參考答案
A-B,連 L2 一起留。實體鍵相同,A/B 也仍屬 L1。