擅长:python、mysql、java
<p>你真的不需要第二个循环。。只要检查单词[i]是否在字母数组中,就可以使用“includes”或“indexOf”来完成。。你知道吗</p>
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<pre class="snippet-code-js lang-js prettyprint-override"><code>var letter = ['a', 'b', 'c', 'd'];
var word = "Black";
var dictionary = [];
var dictionaryCoincidence = [];
for (var i = 0; i < word.length; i++) {
dictionary.push(word[i]);
if (letter.includes(word[i])) {
dictionaryCoincidence.push(dictionary[i]);
}
}
console.log(dictionary); console.log(dictionaryCoincidence);</code></pre>
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