算法数组Java c++ 的相关内容

【算法学习】1470. 重新排列数组（java / c / c++ / python / go / rust）

1470. 重新排列数组：给你一个数组 nums ，数组中有 2n 个元素，按 [x1,x2,...,xn,y1,y2,...,yn] 的格式排列。请你将数组按 [x1,y1,x2,y2,...,xn,yn] 格式重新排列，返回重排后的数组。样例 1输入： nums = [2,5,1,3,4,7],...

【算法学习】1486. 数组异或操作（java / c / c++ / python / go / rust）

1486. 数组异或操作：给你两个整数，n 和 start 。数组 nums 定义为：nums[i] = start + 2 * i（下标从 0 开始）且 n == nums.length 。请返回 nums 中所有元素按位异或（XOR）后得到的结果。样例 1输入： n = 5, start = 0...

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【算法】1720. 解码异或后的数组（java / c / c++ / python / go / rust）

1720. 解码异或后的数组：未知整数数组 arr 由 n 个非负整数组成。经编码后变为长度为 n - 1 的另一个整数数组 encoded ，其中 encoded[i] = arr[i] XOR arr[i + 1] 。例如，arr = [1,0,2,1] 经编码后得到 encoded = [1...

【算法】1480. 一维数组的动态和（java / c / c++ / python / go / rust）

1480. 一维数组的动态和：给你一个数组 nums 。数组「动态和」的计算公式为：runningSum[i] = sum(nums[0]…nums[i]) 。请返回 nums 的动态和。样例 1输入： nums = [1,2,3,4] 输出： [1,3,6,10] 解释：动态和计算过程为 [1,...

【算法】1929. 数组串联（java / c / c++ / python / go / rust）

1929. 数组串联：给你一个长度为 n 的整数数组 nums 。请你构建一个长度为 2n 的答案数组 ans ，数组下标从 0 开始计数，对于所有 0 <= i < n 的 i ，满足下述所有要求：ans[i] == nums[i]ans[i + n] == nums[i]具体而言...

【算法】1920. 基于排列构建数组（java / c / c++ / python / go / rust）

1920. 基于排列构建数组：给你一个从 0 开始的排列 nums（下标也从 0 开始）。请你构建一个同样长度的数组 ans ，其中，对于每个 i（0 <= i < nums.length），都满足 ans[i] = nums[nums[i]] 。返回构建好的数组 ans 。从 0...

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