# Product of Array Exclude Itself### Source- lintcode: [(50) Product of Array Exclude Itself](http://www.lintcode.com/en/problem/product-of-array-exclude-itself/)- GeeksforGeeks: [A Product Array Puzzle - GeeksforGeeks](http://www.geeksforgeeks.org/a-product-array-puzzle/)~~~Given an integers array A.Define B[i] = A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1], calculate B WITHOUT divide operation.ExampleFor A=[1, 2, 3], return [6, 3, 2].~~~### 题解1 - 左右分治根据题意，有 result[i]=left[i]⋅right[i]result[i] = left[i] \cdot right[i]result[i]=left[i]⋅right[i], 其中 left[i]=∏j=0i−1A[j]left[i] = \prod _{j = 0} ^{i - 1} A[j]left[i]=∏j=0i−1A[j], right[i]=∏j=i+1n−1A[j]right[i] = \prod _{j = i + 1} ^{n - 1} A[j]right[i]=∏j=i+1n−1A[j]. 即将最后的乘积分为两部分求解，首先求得左半部分的值，然后求得右半部分的值。最后将左右两半部分乘起来即为解。### C++~~~class Solution {public: /** * @param A: Given an integers array A * @return: A long long array B and B[i]= A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1] */ vector<long long> productExcludeItself(vector<int> &nums) { const int nums_size = nums.size(); vector<long long> result(nums_size, 1); if (nums.empty() || nums_size == 1) { return result; } vector<long long> left(nums_size, 1); vector<long long> right(nums_size, 1); for (int i = 1; i != nums_size; ++i) { left[i] = left[i - 1] * nums[i - 1]; right[nums_size - i - 1] = right[nums_size - i] * nums[nums_size - i]; } for (int i = 0; i != nums_size; ++i) { result[i] = left[i] * right[i]; } return result; }};~~~### 源码分析一次`for`循环求出左右部分的连乘积，下标的确定可使用简单例子辅助分析。### 复杂度分析两次`for`循环，时间复杂度 O(n)O(n)O(n). 使用了左右两半部分辅助空间，空间复杂度 O(2n)O(2n)O(2n).### 题解2 - 原地求积题解1中使用了左右两个辅助数组，但是仔细瞅瞅其实可以发现完全可以在最终返回结果`result`基础上原地计算左右两半部分的积。### C++~~~class Solution {public: /** * @param A: Given an integers array A * @return: A long long array B and B[i]= A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1] */ vector<long long> productExcludeItself(vector<int> &nums) { const int nums_size = nums.size(); vector<long long> result(nums_size, 1); // solve the left part first for (int i = 1; i < nums_size; ++i) { result[i] = result[i - 1] * nums[i - 1]; } // solve the right part long long temp = 1; for (int i = nums_size - 1; i >= 0; --i) { result[i] *= temp; temp *= nums[i]; } return result; }};~~~### 源码分析计算左半部分的递推式不用改，计算右半部分的乘积时由于会有左半部分值的干扰，故使用`temp`保存连乘的值。注意`temp`需要使用`long long`, 否则会溢出。### 复杂度分析时间复杂度同上，空间复杂度为 O(1)O(1)O(1).
