javascript 如果您不知道javascript中每个数组的长度，如何比较两个不同长度的数组？
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how to compare two arrays of different length if you dont know the length of each one in javascript?
提问by bentham
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).
我被困在这。我有 2 个数组，我不知道每个数组的长度，它们可以是相同的长度，也可以不是，我不知道，但我需要创建一个新数组，其中的数字在 (2, 10）。
For this case:
对于这种情况：
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{}
I don't know why my code never reach the first condition and I don't know what to do when b has more length than a.
我不知道为什么我的代码永远不会达到第一个条件，而且我不知道当 b 的长度大于 a 时该怎么办。
回答by BudgieInWA
It seems that you have a logic error in your code, if I am understanding your requirements correctly.
如果我正确理解您的要求，您的代码中似乎存在逻辑错误。
This code will put all elements that are in athat are not in b, into newArray.
此代码会将所有a不在 , 中的元素b放入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
澄清一下，如果
a = [2, 4, 10];
b = [4, 3, 11, 12];
then newArraywill be [2,10]
然后newArray将是[2,10]
回答by Selvakumar Ponnusamy
Try this
试试这个
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;
}
