如何高效筛选sortKey中与receivedOrderKey部分匹配的元素？

我有两个数组：

const sortKey = ["invoiceDate-desc", "invoiceDate-asc", "location-asc", "location-desc", "orderId-asc", "orderId-desc", "invoiceId-asc", "invoiceId-desc", "type-asc", "type-desc", "total-asc", "total-desc"];
const receivedOrderKey = ["invoiceId", "orderId"];

我需要筛选出sortKey中所有包含receivedOrderKey里元素的值，最终结果应该是["invoiceId-desc", "invoiceId-asc", "orderId-asc", "orderId-desc"]。

目前我用双重for循环实现了需求：

const requestedOptions = [];
for(i=0;i<sortKey.length;i++){
    var str1 = sortKey[i].toLowerCase();
    for(j=0;j<receivedOrderKey.length;j++){
        var str2 = receivedOrderKey[j].toLowerCase();
        if(str1.includes(str2)) {
            requestedOptions.push(sortKey[i]);
        }
    }
}

但这种方式的时间复杂度是O(n*m)，有没有更高效的实现方式？

更高效的实现思路 #

双重循环的问题在于每次检查都要遍历receivedOrderKey，当数组规模变大时效率会明显下降。我们可以通过将匹配关键词转为查找效率更高的数据结构，或者利用正则匹配来优化，把时间复杂度降到O(n)级别。

方法1：Set + filter + some（通用性强）

把receivedOrderKey转换成Set（查找时间复杂度O(1)），再结合数组的filter和some方法实现筛选：

const sortKey = ["invoiceDate-desc", "invoiceDate-asc", "location-asc", "location-desc", "orderId-asc", "orderId-desc", "invoiceId-asc", "invoiceId-desc", "type-asc", "type-desc", "total-asc", "total-desc"];
const receivedOrderKey = ["invoiceId", "orderId"];

// 提前把关键词转成小写并存入Set，避免重复转换
const lowerKeySet = new Set(receivedOrderKey.map(key => key.toLowerCase()));

const requestedOptions = sortKey.filter(item => {
    const lowerItem = item.toLowerCase();
    // 检查当前item是否包含Set中的任意一个关键词
    return [...lowerKeySet].some(key => lowerItem.includes(key));
});

console.log(requestedOptions); 
// 输出: ["orderId-asc", "orderId-desc", "invoiceId-asc", "invoiceId-desc"]

方法2：正则表达式（适合关键词无特殊字符场景）

如果receivedOrderKey里的元素都是普通字符串（没有正则特殊字符，比如*、+等），可以用正则表达式简化代码：

const sortKey = ["invoiceDate-desc", "invoiceDate-asc", "location-asc", "location-desc", "orderId-asc", "orderId-desc", "invoiceId-asc", "invoiceId-desc", "type-asc", "type-desc", "total-asc", "total-desc"];
const receivedOrderKey = ["invoiceId", "orderId"];

// 生成匹配任意关键词的正则，开启忽略大小写模式
const matchRegex = new RegExp(receivedOrderKey.join('|'), 'i');

const requestedOptions = sortKey.filter(item => matchRegex.test(item));

console.log(requestedOptions);
// 输出同样的目标结果

优化点说明 #

• Set方法：将receivedOrderKey转为Set后，每次关键词查找的时间从O(m)降到O(1)，整体筛选过程的时间复杂度为O(n)（n为sortKey的长度），数组规模越大，效率提升越明显。
• 正则方法：正则引擎内部经过高度优化，这种简单的"或匹配"效率很高，同时代码量更少、可读性更强。
• 另外你原来的代码存在全局变量污染问题（i和j未用let/const声明），上面的优化方案也修复了这个问题。

内容的提问来源于stack exchange，提问作者Nightshade