lc 238 Product of Array Except Self 238 Product of Array Except Self Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i]. Solve it without divis…

剑指offer 66. 构建乘积数组 题目: 给定一个数组A[0, 1, ..., n-1],请构建一个数组B[0, 1, ..., n-1],其中B中的元素B[i] = A[0] * A[1] * ... * A[i-1] * A[i +1] ... A[n-1].不能使用除法. 同leetcode 238 https://leetcode.com/problems/product-of-array-except-self/ 分析: 假如可以利用除法,则利用(A[0]*A[1]*...A[n-…

Given an array nums of n integers where n > 1,  return an array output such that output[i] is equal to the product of all the elements of numsexcept nums[i]. Example: Input: [1,2,3,4] Output: [24,12,8,6] Note: Please solve it without division and in…

Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i]. Solve it without division and in O(n). For example, given [1,2,3,4], return [24,12,8,6]. Fo…

Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i]. Solve it without division and in O(n). For example, given [1,2,3,4], return [24,12,8,6]. Fo…

题目描述: Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i]. Solve it without division and in O(n). For example, given [1,2,3,4], return [24,12,8,…

Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i]. Solve it without division and in O(n). For example, given [1,2,3,4], return [24,12,8,6]. Fo…

Given an array nums of n integers where n > 1,  return an array output such that output[i] is equal to the product of all the elements of nums except nums[i]. Example: Input: [1,2,3,4] Output: [24,12,8,6] Note: Please solve it without division and in…

这题看似简单,不过两个要求很有意思: 1.不准用除法:最开始我想到的做法是全部乘起来,一项项除,可是中间要是有个0,这做法死得很惨. 2.空间复杂度O(1):题目说明了返回的那个数组不算进复杂度分析里面 做法:既然不用除法,对于某个数i, result[i] = 0到i - 1的乘积 X   i + 1 到 n - 1的乘积 具体来说,先正向遍历,result[i] 存储的是 0 到i - 1的乘积,再反向遍历,乘上另一半,这就同时满足了时间复杂度和空间复杂度的要求 class Solution…

原题 思路: 注意时间复杂度,分别乘积左右两边,可达到O(n) class Solution { public: vector<int> productExceptSelf(vector<int> &nums) { int len = nums.size(); vector<int> res(len, 1); int left = 1, right = 1; for (int j = 1; j < len; j++) { left *= nums[j -…