比较 Javascript 中的两个数组 - 返回差异
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原文地址: http://stackoverflow.com/questions/3432929/
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Comparing two arrays in Javascript - Returning differences
提问by Hans
Assuming we have:
假设我们有：
array1 = ['A', 'B', 'C', 'D', 'E']; array2 = ['C', 'E'];
Is there a proven and fast solution to compare two arrays against each other, returning one array without the values appearing in both arrays (C and E here). So:
是否有一种行之有效的快速解决方案来比较两个数组，返回一个数组而没有出现在两个数组中的值（此处为 C 和 E）。所以：
array3 = ['A', 'B', 'D']
should be the output of the solution. (jquery may be involved)
应该是解决方案的输出。（jquery可能涉及）
thx.
谢谢。
回答by Hans
I accepted Matthews Solution, but dont want to ignore a different faster solution i just found.
我接受了 Matthews 解决方案，但不想忽略我刚刚找到的另一个更快的解决方案。
var list1 = [1, 2, 3, 4, 5, 6];
var list2 = ['a', 'b', 'c', 3, 'd', 'e'];
var lookup = {};
for (var j in list2) {
lookup[list2[j]] = list2[j];
}
for (var i in list1) {
if (typeof lookup[list1[i]] != 'undefined') {
alert('found ' + list1[i] + ' in both lists');
break;
}
}
Source: Optimize Loops to Compare Two Arrays
来源：优化循环以比较两个数组
回答by Matthew Flaschen
This is a set difference. A simple implementation is:
这是一个集合差异。一个简单的实现是：
jQuery.grep(array1, function(el)
{
return jQuery.inArray(el, array2) == -1;
});
This is O(m * n), where those are the sizes of the arrays. You cando it in O(m + n), but you need to use some kind of hash set. You can use a JavaScript object as a simple hash set for strings. For relatively small arrays, the above should be fine.
这是 O(m * n)，其中这些是数组的大小。您可以在 O(m + n) 中完成，但您需要使用某种散列集。您可以将 JavaScript 对象用作字符串的简单哈希集。对于相对较小的数组，以上应该没问题。
回答by akonsu
a proven fast solution that i know of is a binary search that you can use after you sort one of the arrays. so the solution takes time that depends on the sorting algorithm. but is at least log(N).
我所知道的一个经过验证的快速解决方案是二分搜索，您可以在对其中一个数组进行排序后使用它。所以解决方案需要时间，这取决于排序算法。但至少是 log(N)。
