如何比较两个字符串数组,不区分大小写和独立于排序 - JS,ES6

Question

我想比较两个字符串数组,但不区分大小写且独立.

例如:

['a', 'b', 'c'] === ['A', 'c', 'B'] -> TRUE

['a', 'b', 'c'] === ['a', 'b', 'd'] -> FALSE

TRUE当它们具有相同的长度和相同的值(不区分大小写['A'] === ['a'] -> true)且独立时,关于排序['a', 'b'] === ['b', 'a'] -> true.

我现在做的是:

areEqual = (arr1, arr2) => {
    const equalLength = arr1.length === arr2.length;

    return arr2.every(arr2Item => {

        return arr1.includes(arr2Item.toLowerCase());

    }) && equalLength;
};

,但这是case sensitive.

我使用JS, ES6带React.

Answer 1

您可以将字符串规范化为小写,并使用a Set来检查值.

function compare(a, b) {
    const lower = s => s.toLowerCase();
    return b
        .map(lower)
        .every(Set.prototype.has, new Set(a.map(lower)));
}

console.log(compare(['a', 'b', 'c'], ['A', 'c', 'B'])); //  true
console.log(compare(['a', 'b', 'c'], ['a', 'b', 'd'])); // false
console.log(compare(['a', 'b', 'c'], ['a', 'c']));      //  true