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
如果你的数组是排序的,这应该运行在O(n),其中n是min(a.length, b.length)
function intersect_1d( a, b ){
var out=[], ai=0, bi=0, acurr, bcurr, last=Number.MIN_SAFE_INTEGER;
while( ( acurr=a[ai] )!==undefined && ( bcurr=b[bi] )!==undefined ){
if( acurr < bcurr){
if( last===acurr ){
out.push( acurr );
}
last=acurr;
ai++;
}
else if( acurr > bcurr){
if( last===bcurr ){
out.push( bcurr );
}
last=bcurr;
bi++;
}
else {
out.push( acurr );
last=acurr;
ai++;
bi++;
}
}
return out;
}
其他回答
Var数组= [ [1,2,3], [2,3,4,5] ] 函数commonValue(…arr) { 令res = arr[0]。过滤(函数(x) { 返回arr.every((y) => y.includes(x)) }) 返回res; } commonValue数组(…);
希望这有助于所有版本。
function diffArray(arr1, arr2) {
var newArr = [];
var large = arr1.length>=arr2.length?arr1:arr2;
var small = JSON.stringify(large) == JSON.stringify(arr1)?arr2:arr1;
for(var i=0;i<large.length;i++){
var copyExists = false;
for(var j =0;j<small.length;j++){
if(large[i]==small[j]){
copyExists= true;
break;
}
}
if(!copyExists)
{
newArr.push(large[i]);
}
}
for(var i=0;i<small.length;i++){
var copyExists = false;
for(var j =0;j<large.length;j++){
if(large[j]==small[i]){
copyExists= true;
break;
}
}
if(!copyExists)
{
newArr.push(small[i]);
}
}
return newArr;
}
另一种可以同时处理任意数量数组的索引方法:
// Calculate intersection of multiple array or object values.
function intersect (arrList) {
var arrLength = Object.keys(arrList).length;
// (Also accepts regular objects as input)
var index = {};
for (var i in arrList) {
for (var j in arrList[i]) {
var v = arrList[i][j];
if (index[v] === undefined) index[v] = 0;
index[v]++;
};
};
var retv = [];
for (var i in index) {
if (index[i] == arrLength) retv.push(i);
};
return retv;
};
它只适用于可以作为字符串计算的值,你应该将它们作为一个数组传递:
intersect ([arr1, arr2, arr3...]);
...但它透明地接受对象作为参数或任何要交叉的元素(总是返回公共值的数组)。例子:
intersect ({foo: [1, 2, 3, 4], bar: {a: 2, j:4}}); // [2, 4]
intersect ([{x: "hello", y: "world"}, ["hello", "user"]]); // ["hello"]
编辑:我只是注意到,这是,在某种程度上,有点bug。
也就是说:我在编码时认为输入数组本身不能包含重复(正如所提供的示例那样)。
但如果输入数组恰好包含重复,就会产生错误的结果。示例(使用下面的实现):
intersect ([[1, 3, 4, 6, 3], [1, 8, 99]]);
// Expected: [ '1' ]
// Actual: [ '1', '3' ]
幸运的是,这很容易通过添加二级索引来解决。那就是:
变化:
if (index[v] === undefined) index[v] = 0;
index[v]++;
by:
if (index[v] === undefined) index[v] = {};
index[v][i] = true; // Mark as present in i input.
,:
if (index[i] == arrLength) retv.push(i);
by:
if (Object.keys(index[i]).length == arrLength) retv.push(i);
完整的例子:
// Calculate intersection of multiple array or object values.
function intersect (arrList) {
var arrLength = Object.keys(arrList).length;
// (Also accepts regular objects as input)
var index = {};
for (var i in arrList) {
for (var j in arrList[i]) {
var v = arrList[i][j];
if (index[v] === undefined) index[v] = {};
index[v][i] = true; // Mark as present in i input.
};
};
var retv = [];
for (var i in index) {
if (Object.keys(index[i]).length == arrLength) retv.push(i);
};
return retv;
};
intersect ([[1, 3, 4, 6, 3], [1, 8, 99]]); // [ '1' ]
对于只包含字符串或数字的数组,你可以做一些排序,就像其他答案一样。对于任意对象数组的一般情况,我不认为你可以避免这样做。下面将为您提供任意数量的数组的交集作为arrayIntersection的参数:
var arrayContains = Array.prototype.indexOf ?
function(arr, val) {
return arr.indexOf(val) > -1;
} :
function(arr, val) {
var i = arr.length;
while (i--) {
if (arr[i] === val) {
return true;
}
}
return false;
};
function arrayIntersection() {
var val, arrayCount, firstArray, i, j, intersection = [], missing;
var arrays = Array.prototype.slice.call(arguments); // Convert arguments into a real array
// Search for common values
firstArray = arrays.pop();
if (firstArray) {
j = firstArray.length;
arrayCount = arrays.length;
while (j--) {
val = firstArray[j];
missing = false;
// Check val is present in each remaining array
i = arrayCount;
while (!missing && i--) {
if ( !arrayContains(arrays[i], val) ) {
missing = true;
}
}
if (!missing) {
intersection.push(val);
}
}
}
return intersection;
}
arrayIntersection( [1, 2, 3, "a"], [1, "a", 2], ["a", 1] ); // Gives [1, "a"];
这是一个提议的标准:对于当前阶段2的提议https://github.com/tc39/proposal-set-methods,您可以使用
mySet.intersection(mySet2);
在此之前,你可以使用Immutable.js的Set,它激发了这个提议
Immutable.Set(mySet).intersect(mySet2)