关键词:lodash.flattenDeep 实现 以下是用 JavaScript 手写实现类似于 lodash.flattenDeep 的函数来将数组递归展平为一维数组: function flattenDeep(arr) { let result = []; for (let item of arr) { if (Array.isArray(item)) { result = result.concat(flattenDeep(it
It's still an Amazon interview question. Given an array containing only stars '*' and hashes '#' . Find longest contiguous sub array that will contain equal no. of stars '*' and hashes '#'.Output the index range if the longest contiguous sub array does exist or else output -1 and ...
Vba - Naming an array using a variable, 1 You can create an array of arrays (though your question is a little unclear..) Sub MyArrays () Dim arrays () Dim arr Dim i, j i = 5 'e.g. ReDim arrays (1 To i) For j = 1 To i arr = Array () ReDim arr (1 To j) arrays ...
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Subscribeto see which companies asked this question Show Tags Show Similar Problems Discuss 求给定数组的第K大的元素。 假设先排序,然后取第K个元素,那么时间复杂度是O(n*log n)。 借助堆的数据结构,能够把时间复杂度降到O(n*logk)。 假设求第K大的元素,那么要构建的是小顶堆。
for(inti = 0; i < num.Length; i++) { for(intj = i + 1; j < num.Length; j++) if(num[i] > num[j]) { t = num[i]; num[i] = num[j]; num[j] = t; } } Console.WriteLine("Sorting elements in Ascending Order :\n"); ...
判断数组的方法,并对比它们的优劣 Diamond2128/fe-interview#58 Open DennisYongLe commented Sep 15, 2023 这里在理解的时候可能会把Object.prototype.toString.call()与[12].toString()方法相混淆,如果把这两个toString方法当作是一个,那么后续的理解会造成困难。因为当我们期待Object.prototype.toString.call([12...
Also, 3 is the first index where this occurs. Example 2: Input: nums = [1, 2, 3] Output: -1 Explanation: There is no index that satisfies the conditions in the problem statement. Note: The length of nums will be in the range [0, 10000]. ...
0 4.1k 1 Post Your Answer Jun, 201516 http://www.dotnetfunda.com/interviews/show/5467/how-to-assign-a-computed-value-to-the-same-array-back 0 Whats the work of Let Operator in LINQ? Select and Delete Data Using LINQ With WCF?
My code: publicintgetMinLength(int[]nums){if(nums==null||nums.length==0){return0;}intmin=Integer.MAX_VALUE;intmax=Integer.MIN_VALUE;for(inti=1;i<nums.length;i++){if(nums[i]>nums[i-1]){continue;}else{min=Math.min(min,nums[i]);max=Math.max(max,nums[i-1]);}}if(max==Inte...