array_n_element = sorted_array.length - 1; // Get the product of three largest integers in sorted array for (var x = array_n_element; x > array_n_element - 3; x--) { product1 = product1 * sorted_array[x]; } product2 = sorted_array[0] * sorted_array[1] * sorted_array[a...
function radixSort(array) { let maxDigitCount = maxDigits(array); for(let i = 0; i < maxDigitCount; i++) { let digitBucket = Array.from({length: 10}, () => []); for(let j = 0; j < array.length; j++) { let lastDigit = getDigit(array[j], i); digitBucket[lastDigit]...
functionoddArray(arr){letresult=[];functionhelperRecursiveFn(arr){if(arr.length===0){return;// 1}elseif(arr[0]%2!==0){result.push(arr[0]);// 2}helperRecursiveFn(arr.slice(1));// 3}helperRecursiveFn(arr);returnresult;}oddArray([1,2,3,4,5,6,7,8,9,10]);// OutPut -> [...
Example:Printing the elements in an array O(n2) - Quadratic Quadratic time complexity is almost the inverse of logarithmic complexity. With Quadratic Complexity execution time increases at an increasing rate. Quadratic time suggests that the function’s run time is proportional to the square of the...
function sortIntegers(a, b) { return a - b; } // greatest product is either (min1 * min2 * max1 || max1 * max2 * max3) function computeProduct(unsorted) { var sorted_array = unsorted.sort(sortIntegers), product1 = 1,
其实ES5中已经包含了Array类,ES6中已经包含了Set类,可直接使用 代码实现一个Set类 function Set() { this.item = {} Set.prototype.add = function (value) { if (this.has(value)) { return false } this.item[value] = value return true } Set.prototype.has = function (value) { return this.it...
因此,算法的time-complexity将是O(2N)=O(N) 我的排序算法的时间复杂度是多少? 是的,正如你在问题中所说,这是一个二次排序算法。原因是: 代码的主要部分运行嵌套循环,如下所示: for(int i = 1; i < n; i++) { int j = i-1 while (j >= 0... 在内部循环中不断地工作。 在最坏的情况下,...
unsorted.sort(sortIntegers), product1 = 1, product2 = 1, array_n_element = sorted_array.length - 1; // Get the product of three largest integers in sorted array for (var x = array_n_element; x > array_n_element - 3; x--) { product1 = product1 * sorted_array[x]; } ...
function sortIntegers(a, b) {return a - b;} // greatest product is either (min1 * min2 * max1 || max1 * max2 * max3)function computeProduct(unsorted) {var sorted_array = unsorted.sort(sortIntegers),product1 = 1,product2 = 1,array...
@param array 待排序数组*/ function heapSort(array) { console.time('堆排序耗时'); if (Object.prototype.toString.call(array).slice(8, -1) === 'Array') { //建堆 var heapSize = array.length, temp; for (var I = Math.floor(heapSize / 2) - 1; I >= 0; I—) { ...