給定兩個(gè)長度相同的整數(shù)數(shù)組，將其中的數(shù)字分別一一配對，對每一對數(shù)字計(jì)算乘積，然后求和，計(jì)算出總和最小的配對方式，并打印出總和

let arr1 = [1, 2, 3, 4];
let arr2 = [1, 2, 3, 4];

arr1 = arr1.sort((a, b) => a - b);
arr2 = arr2.sort((a, b) => b - a);

let min = 0;
for (let i = 0; i < arr1.length; i++) {
    min += arr1[i] * arr2[i]
}

console.log(min);

應(yīng)該不用一一匹對吧