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January 23, 2017 00:43
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Find a product within an array of numbers. Demo at http://jsbin.com/zidogenaro/edit?js,console
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| var mathManager = { | |
| findProduct: function(product, arr) { | |
| var result = null; | |
| for (var i=0; i<arr.length; i++) { | |
| for (var j=i+1; j<arr.length; j++) { | |
| if (product === arr[i] * arr[j]) { | |
| result = [arr[i], arr[j]]; | |
| break; | |
| } | |
| } | |
| if (result) { | |
| break; | |
| } | |
| } | |
| return result; | |
| } | |
| }; | |
| var arr = [1, 3, 6, 8, 9, 2, 0, 11, 7]; | |
| var result = mathManager.findProduct(18, arr); | |
| console.log('Find product 18 in ' + arr); | |
| console.log(result); | |
| console.log(result.length===2 && result.every(function(v,i) { return v === [3,6][i]; })); | |
| result = mathManager.findProduct(19, arr); | |
| console.log('Find product 19 in ' + arr); | |
| console.log(result); | |
| console.log(result === null); | |
| var result = mathManager.findProduct(22, arr); | |
| console.log('Find product 22 in ' + arr); | |
| console.log(result); | |
| console.log(result.length===2 && result.every(function(v,i) { return v === [2,11][i]; })); |
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