javascript 求 N 个数组的笛卡尔积

[sunhengzhe]

给定 N 个数组，求出这些数组的笛卡尔积

function getArranging( /* listA, listB, ... */){
      //TODO - 返回 listA × listB × ...
}

const a = [0, 1];
const b = [3, 4, 5];
const c = [6, 7];

// test
expect(getArranging( a, b )).to.eql(
    [
        [0, 3], [0, 4], [0, 5],
        [1, 3], [1, 4], [1, 5]
    ]
);

expect(getArranging( a, b, c )).to.eql(
    [
        [0,3,6], [0,3,7],
        [0,4,6], [0,4,7],
        [0,5,6], [0,5,7],
        [1,3,6], [1,3,7],
        [1,4,6], [1,4,7],
        [1,5,6], [1,5,7]
    ]
);

[yinkaihui]

function getArranging(...args) {
  return args.reduce(function (a, b) {
    const curRes = []
    a.forEach(x => {
      b.forEach(y => {
        curRes.push([].concat(x, y))
      })
    })
    return curRes
  }, [[]])
}

[iLove-Coding]

       function getArranging() {
            let result = arguments[0];
            for (let i = 1; i < arguments.length; i++) {
                result = utilFn(result, arguments[i]);
            }
            return result;
        }

        function utilFn(arr1, arr2) {
            let arr = [];
            arr1.map(elt1 => {
                arr2.map(elt2 => {
                    arr.push([].concat(elt1, elt2));
                })
            });
            return arr;
        }

[sunhengzhe]

😂 我的思路跟 @yinkaihui 是一样的，另外补充一个递归的写法，盗用一下 @iLove-Coding 的 utilFn

function getArranging(listA, listB, ...rest) {
    if (!listB) {
        return listA;
    }

    return getArranging(utilFn(listA, listB), ...rest);
}