* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
1.7 KiB
1.7 KiB
id, title, challengeType, videoUrl, forumTopicId, dashedName
id | title | challengeType | videoUrl | forumTopicId | dashedName |
---|---|---|---|---|---|
56533eb9ac21ba0edf2244e1 | 循环嵌套 | 1 | https://scrimba.com/c/cRn6GHM | 18248 | nesting-for-loops |
--description--
如果你有一个二维数组,可以使用相同的逻辑,先遍历外面的数组,再遍历里面的子数组。下面是一个例子:
var arr = [
[1,2], [3,4], [5,6]
];
for (var i=0; i < arr.length; i++) {
for (var j=0; j < arr[i].length; j++) {
console.log(arr[i][j]);
}
}
一次输出arr
中的每个子元素。提示,对于内部循环,我们可以通过arr[i]
的.length
来获得子数组的长度,因为arr[i]
的本身就是一个数组。
--instructions--
修改函数multiplyAll
,获得arr
内部数组的每个数字相乘的结果product
。
--hints--
multiplyAll([[1],[2],[3]])
应该返回 6
。
assert(multiplyAll([[1], [2], [3]]) === 6);
multiplyAll([[1,2],[3,4],[5,6,7]])
应该返回 5040
。
assert(
multiplyAll([
[1, 2],
[3, 4],
[5, 6, 7]
]) === 5040
);
multiplyAll([[5,1],[0.2, 4, 0.5],[3, 9]])
应该返回 54
。
assert(
multiplyAll([
[5, 1],
[0.2, 4, 0.5],
[3, 9]
]) === 54
);
--seed--
--seed-contents--
function multiplyAll(arr) {
var product = 1;
// Only change code below this line
// Only change code above this line
return product;
}
multiplyAll([[1,2],[3,4],[5,6,7]]);
--solutions--
function multiplyAll(arr) {
var product = 1;
for (var i = 0; i < arr.length; i++) {
for (var j = 0; j < arr[i].length; j++) {
product *= arr[i][j];
}
}
return product;
}
multiplyAll([[1,2],[3,4],[5,6,7]]);