I am stuck in this. I got 2 arrays, I don't know the length of each one, they can be the same length or no, I don't know, but I need to create a new array with the numbers no common in just a (2, 10).
var a = [2,4,10];
var b = [1,4];
var newArray = [];
if(a.length >= b.length ){
for(var i =0; i < a.length; i++){
for(var j =0; j < b.length; j++){
if(a[i] !=b [j]){
newArray.push(b);
}
}
}
}else{}
It seems that you have a logic error in your code, if I am understanding your requirements correctly.
This code will put all elements that are in a
that are not in b
, into newArray
.
var a = [2, 4, 10];
var b = [1, 4];
var newArray = [];
for (var i = 0; i < a.length; i++) {
// we want to know if a[i] is found in b
var match = false; // we haven't found it yet
for (var j = 0; j < b.length; j++) {
if (a[i] == b[j]) {
// we have found a[i] in b, so we can stop searching
match = true;
break;
}
// if we never find a[i] in b, the for loop will simply end,
// and match will remain false
}
// add a[i] to newArray only if we didn't find a match.
if (!match) {
newArray.push(a[i]);
}
}
To clarify, if
a = [2, 4, 10];
b = [4, 3, 11, 12];
then newArray
will be [2,10]
Try this
var a = [2,4,10];
var b = [1,4];
var nonCommonArray = [];
for(var i=0;i<a.length;i++){
if(!eleContainsInArray(b,a[i])){
nonCommonArray.push(a[i]);
}
}
function eleContainsInArray(arr,element){
if(arr != null && arr.length >0){
for(var i=0;i<arr.length;i++){
if(arr[i] == element)
return true;
}
}
return false;
}
I found this solution just using the filter()<\/code> and
include()<\/code> methods, a very and short easy one.
function compareArrays(a, b) {
return a.filter(e => b.includes(e));
}
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.