查找字符串中的字符是否与字符串数组中的字符匹配

Question

我正在尝试实现这样的目标：

const word = "abracadabra" 让 usedCharacter = ["a", "b", "c", "d", "r"]

如果单词由可用的 usedCharacter 组成，则返回 true

让 usedCharacter = ["a", "b", "c", "d", "r", "z"] 也应该返回 true

让 usedCharacter = ["b", "c", "d", "r", "z"] 应该返回false

我应该使用 for 循环遍历 word 并将每个 word(i) 与使用的字符进行比较还是有更简单的方法？

感谢您的意见！

Answer 1

你可以这样做：

const word = "abracadabra";
let usedCharacter = ["a", "b", "c", "d", "r"];

let res = word.split('').filter(function(char) {
    return !usedCharacter.includes(char);
});
if(res.length == 0) {
     return true;
} else {
     return false;
}

这样你就可以知道哪些字符丢失了

祝你好运

Answer 2

您可以使用splitting 与每个字符一起使用单词轻松实现此目的，并检查usedCharacter 中是否存在每个字符。

const word = "abracadabra";
const usedCharacter = ["a", "b", "c", "d", "r"];

function hasAllUsedCharacters(word, used) {
  return word.split("").every((c) => usedCharacter.includes(c));
}

const result = hasAllUsedCharacters(word, usedCharacter);
console.log(result);

使用Set的替代解决方案

const word = "abracadabra";
const usedCharacter = ["a", "b", "c", "d", "r"];

function hasAllUsedCharacters(word, used) {
  const set = new Set(usedCharacter);
  const initialLength = set.size;
  word.split("").forEach((w) => set.add(w));

  return set.size === initialLength;
}

const result = hasAllUsedCharacters(word, usedCharacter);
console.log(result);