Javascript 多阵列推送错误 let evenOdd=(数组)=>{ 让newArray=[],[]; 对于(i=0;i
您需要删除第二个访问:Javascript 多阵列推送错误 let evenOdd=(数组)=>{ 让newArray=[],[]; 对于(i=0;i,javascript,arrays,Javascript,Arrays,您需要删除第二个访问: let evenOdd = (array) => { let newArray = [[],[]]; for (i=0; i<array.length; i++) { if (array[i] % 2 === 0) { newArray[0][0].push(array[i]); } else { newArray[0][1].push(array[i]); } } return newArra
let evenOdd = (array) => {
let newArray = [[],[]];
for (i=0; i<array.length; i++) {
if (array[i] % 2 === 0) {
newArray[0][0].push(array[i]);
} else {
newArray[0][1].push(array[i]);
}
}
return newArray;
}
let evenOdd=(数组)=>{
让newArray=[],[];
对于(i=0;i无需为内部数组指定索引,更新代码:
newArray[0][0].push(array[i]);
^
var evenOdd=函数(数组){
var newArray=[],[];
对于(i=0;i你应该这样做
var evenOdd = function(array) {
var newArray = [[], []];
for (i=0; i<array.length; i++) {
if (array[i] % 2 === 0) {
newArray[0].push(array[i]);
} else {
newArray[1].push(array[i]);
}
}
return newArray;
}
console.log(evenOdd([1,2,3,4,5,6]));
很好的方法,但不是最优的,因为它将循环数组两次。
const oddEven = array => [
array.filter(e => e%2),
array.filter(e => !(e%2))
];