用javascript实现数组交叉的最简单、无库代码是什么?我想写
intersection([1,2,3], [2,3,4,5])
并获得
[2, 3]
用javascript实现数组交叉的最简单、无库代码是什么?我想写
intersection([1,2,3], [2,3,4,5])
并获得
[2, 3]
当前回答
使用Underscore.js或lodash.js
_.intersection( [0,345,324] , [1,0,324] ) // gives [0,324]
其他回答
这是我使用的一个非常简单的实现。它是无损的,也确保不复制整个。
Array.prototype.contains = function(elem) {
return(this.indexOf(elem) > -1);
};
Array.prototype.intersect = function( array ) {
// this is naive--could use some optimization
var result = [];
for ( var i = 0; i < this.length; i++ ) {
if ( array.contains(this[i]) && !result.contains(this[i]) )
result.push( this[i] );
}
return result;
}
function intersectionOfArrays(arr1, arr2) {
return arr1.filter((element) => arr2.indexOf(element) !== -1).filter((element, pos, self) => self.indexOf(element) == pos);
}
如果你需要让它处理交叉多个数组:
Const intersect = (a1, a2,…rest) => { Const a12 = a1。过滤器(value => a2.includes(value)) 如果休息。长度=== 0){返回a12;} 回归相交(a12,…rest); }; console.log(相交([1、2、3、4、5],[1,2],[1、2、3、4、5],[2 10 1]))
对这里最小的一个(filter/indexOf解决方案)稍作调整,即使用JavaScript对象在其中一个数组中创建值的索引,将从O(N*M)减少到“可能”线性时间。source1 source2
function intersect(a, b) {
var aa = {};
a.forEach(function(v) { aa[v]=1; });
return b.filter(function(v) { return v in aa; });
}
这不是最简单的解决方案(它的代码比filter+indexOf要多),也不是最快的解决方案(可能比intersect_safe()慢一个常数因子),但似乎是一个很好的平衡。它非常简单,同时提供了良好的性能,并且不需要预先排序的输入。
破坏性似乎是最简单的,特别是如果我们可以假设输入是排序的:
/* destructively finds the intersection of
* two arrays in a simple fashion.
*
* PARAMS
* a - first array, must already be sorted
* b - second array, must already be sorted
*
* NOTES
* State of input arrays is undefined when
* the function returns. They should be
* (prolly) be dumped.
*
* Should have O(n) operations, where n is
* n = MIN(a.length, b.length)
*/
function intersection_destructive(a, b)
{
var result = [];
while( a.length > 0 && b.length > 0 )
{
if (a[0] < b[0] ){ a.shift(); }
else if (a[0] > b[0] ){ b.shift(); }
else /* they're equal */
{
result.push(a.shift());
b.shift();
}
}
return result;
}
非破坏性的要稍微复杂一点,因为我们要跟踪指标:
/* finds the intersection of
* two arrays in a simple fashion.
*
* PARAMS
* a - first array, must already be sorted
* b - second array, must already be sorted
*
* NOTES
*
* Should have O(n) operations, where n is
* n = MIN(a.length(), b.length())
*/
function intersect_safe(a, b)
{
var ai=0, bi=0;
var result = [];
while( ai < a.length && bi < b.length )
{
if (a[ai] < b[bi] ){ ai++; }
else if (a[ai] > b[bi] ){ bi++; }
else /* they're equal */
{
result.push(a[ai]);
ai++;
bi++;
}
}
return result;
}