LeetCode 题解（146）: Product of Array Except Self

题目：

Given an array of n integers where n > 1, nums, return an arrayoutput such thatoutput[i] is equal to the product of all the elements ofnums exceptnums[i].

Solve it without division and in O(n).

For example, given [1,2,3,4], return [24,12,8,6].

Follow up:
Could you solve it with constant space complexity? (Note: The output array does not count as extra space for the purpose of space complexity analysis.)

题解：

两种解法，一种递归求当前元素的前半部分积，并通过递归获得后半部分积，相乘得结果。

解法二，  [1, a1, a1 * a2, a1 * a2 * a3] 乘以 [a4 * a3 * a2, a4 * a3, a4, 1]， 用一个变量保存计算的中间结果。

C++版解法一：

class Solution {public:    vector<int> productExceptSelf(vector<int>& nums) {        compute(nums, 1, 0, nums.size());        return nums;    }        int compute(vector<int>& nums, int front, int index, int N) {        int back = 1;        if(index < N) {            int cur = nums[index];            back = compute(nums, front * cur, index + 1, N);            nums[index] = front * back;            back *= cur;        }        return back;    }};

C++版解法二：

class Solution {public:    vector<int> productExceptSelf(vector<int>& nums) {        vector<int> result(nums.size(), 1);                    int p = nums[0];        for(int i = 1; i < nums.size(); i++) {            result[i] = p;            p *= nums[i];        }                p = nums[nums.size()-1];        for(int i = nums.size() - 2; i >= 0; i--) {            result[i] *= p;            p *= nums[i];        }                return result;    }};

Java版解法二：

public class Solution {    public int[] productExceptSelf(int[] nums) {        int[] result = new int[nums.length];        int p = nums[0];        result[0] = 1;        for(int i = 1; i < nums.length; i++) {            result[i] = p;            p *= nums[i];        }                p = nums[nums.length-1];        for(int i = nums.length - 2; i >= 0; i--) {            result[i] *= p;            p *= nums[i];        }                return result;    }}

Python版解法一：

class Solution:    # @param {integer[]} nums    # @return {integer[]}    def productExceptSelf(self, nums):        self.compute(nums, 1, 0, len(nums))        return nums            def compute(self, nums, front, index, total):        back = 1        if index < total:            cur = nums[index]            back = self.compute(nums, front * cur, index + 1, total)            nums[index] = front * back            back *= cur        return back