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javascript - js 两个数组相减

大家讲道理 2017-04-11 09:44:32

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var arr1 = ["", "100", "120", "125", "125", "130", "130"];

var arr2 = ["", "120", "125", "125", "130"];

请问如何求得两个数组之差？使得新数组为 arr3 = ["100","130"]

或则转化为处理两个字符串的问题：
var p1 = 100, 120, 125, 125, 130, 130;
var p2 = 120, 125, 125, 130;

求得两个字符串之差？使得新字符串为 p3 = 100,130;

补充：要求根据元素(字符)个数相减

大家讲道理

光阴似箭催人老，日月如移越少年。

reply all (6)

伊谢尔伦2017-04-11 09:46:32 6 floor

不说废话，直接上代码

var arr1 = ["", "120", "125", "125", "130", "130", "130", "130", "130"]; var arr2 = ["", "120", "125", "130"]; for (var i = arr1.length - 1; i >= 0; i--) { a = arr1[i]; for (var j = arr2.length - 1; j >= 0; j--) { b = arr2[j]; if (a == b) { arr1.splice(i, 1); arr2.splice(j, 1); break; } } } console.log(arr1); // ["125","130","130","130","130"]

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PHPzhong2017-04-11 09:46:32 5 floor

用map（hashtable）的思想来解决。
思路：将第一个数组存入一个对象，这样就已经去重了；再把第二个数组中的元素当作key从对象中删除。
我大概写下代码：

var m = {}; arr1.forEach(function(al){m[al]=al;}) arr2.forEach(function(bl){delete m[bl];}) console.dir(Object.keys(m))

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阿神2017-04-11 09:46:32 4 floor

var arr1 = ["", "100", "120", "125", "125", "130", "130"]; var arr2 = ["", "120", "125", "125", "130"]; var arr3 = []; arr1.forEach((a)=>{ let c = arr2.findIndex(b=>a == b); if (c > -1) delete arr2[c]; else arr3.push(a); }); console.log(arr3) //["100", "130"]

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