Leetcode_C++之238. Product of Array Except Self（除本身元素之外数组其他元素的积）

题目名称

238. Product of Array Except Self

题目描述

Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i].

The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

You must write an algorithm that runs in O(n) time and without using the division operation.

Example 1:
Input: nums = [1,2,3,4]
Output: [24,12,8,6]

Example 2:
Input: nums = [-1,1,0,-3,3]
Output: [0,0,9,0,0]

Constraints:
2 <= nums.length <= 105
-30 <= nums[i] <= 30
The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

Follow up:
Can you solve the problem in O(1) extra space complexity? (The output array does not count as extra space for space complexity analysis.)

初试思路

1、空间换时间，用数组代替一层for循环。
2、重复利用数组空间,避免额外数组，见注释。

初试代码

// 我的代码1
class Solution {
public:
    vector productExceptSelf(vector& nums) {
        // res[i] = num[0]*...*nums[i-1]*  1  *nums[i+1]*.. *nums[nums.size()-1]
        vector rights(nums.size());
        vector results(nums.size());
        rights[nums.size()-1] = 1;
        for(int i=nums.size()-1; i>0; i--){
            rights[i-1] = rights[i] * nums[i];
        }
        int left = 1;
        for(int i=0; i productExceptSelf(vector& nums) {
        // res[i] = num[0]*...*nums[i-1]*  1  *nums[i+1]*.. *nums[nums.size()-1]
        //vector rights(nums.size());
        vector results(nums.size());
        results[nums.size()-1] = 1; //重复利用results来存放rights
        for(int i=nums.size()-1; i>0; i--){
            results[i-1] = results[i] * nums[i];
        }
        int left = 1;
        for(int i=0; i