【刷题-LeetCode】238. Product of Array Except Self

1. Product of Array Except Self

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]

Constraint: It's guaranteed that the product of the elements of any prefix or suffix of the array (including the whole array) fits in a 32 bit integer.

Note: Please solve it without division and in O(n).

Follow up:

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

解 left-right。计算第i个数左边和右边的乘积

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

优化：在从右往左扫描时，right数组中的数字只用了一次，因此用一个变量R代替即可

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