Dav*_*ave 10 javascript algorithm
我试图找到这样的2个数组的每个排列:
// input
lowerWords = ['one', 'two', 'three' ]
upperWords = [ 'ONE', 'TWO', 'THREE' ]
// output
keywords = {
'one two three': true,
'ONE two three': true,
'ONE TWO three': true,
'ONE TWO THREE': true,
'ONE two THREE': true,
'one TWO three': true,
'one two THREE': true,
'one TWO THREE': true,
}
Run Code Online (Sandbox Code Playgroud)
它应具有3个以上的项目,两个数组的长度始终相同。这是我的代码:
const keywords = {}
const lowerWords = ['one', 'two', 'three' ]
const upperWords = [ 'ONE', 'TWO', 'THREE' ]
const wordCount = lowerWords.length
let currentWord = 0
let currentWords = [...upperWords]
while (currentWord < wordCount) {
currentWords[currentWord] = lowerWords[currentWord]
let keyword = currentWords.join(' ')
keywords[keyword] = true
currentWord++
}
currentWord = 0
currentWords = [...lowerWords]
while (currentWord < wordCount) {
currentWords[currentWord] = upperWords[currentWord]
let keyword = currentWords.join(' ')
keywords[keyword] = true
currentWord++
}
Run Code Online (Sandbox Code Playgroud)
结果缺少一些
ONE TWO THREE: true
ONE TWO three: true
ONE two three: true
one TWO THREE: true
one two THREE: true
one two three: true
Run Code Online (Sandbox Code Playgroud)
Nin*_*olz 10
您可以转置数组以获取对数组,然后获取对的所有组合。
const
transpose = array => array.reduce((r, a) => a.map((v, i) => [...(r[i] || []), v]), []),
combinations = array => array.reduce((a, b) => a.reduce((r, v) => r.concat(b.map(w => [].concat(v, w))), []));
var lowerWords = ['one', 'two', 'three'],
upperWords = ['ONE', 'TWO', 'THREE'],
pairs = transpose([lowerWords, upperWords]),
result = combinations(pairs);
console.log(result);Run Code Online (Sandbox Code Playgroud)
.as-console-wrapper { max-height: 100% !important; top: 0; }Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
555 次 |
| 最近记录: |