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 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.)

 

 

使得数组上的每一位数变为其他所有数的乘积

 

C++：

class Solution {
public:
    vector<int> multiply(const vector<int>& A) {
        int len = A.size() ;
        if (len == 0)
            return vector<int>() ;
        vector<int> B(len,1) ;
        int left = 1 ;
        for(int i = 0 ; i < len ; i++){
            B[i] *= left ;
            left *= A[i] ;
        }
        int right = 1 ;
        for(int i = len-1 ; i >= 0 ; i--){
            B[i] *= right ;
            right *= A[i] ;
        }
        return B ;
    }
};

 

 

 

C++（89ms）：

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