[Leetcode] 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.)

class Solution {
public:
/*algorithm
1) for each element A[i], compute two numbers
f[i]: equal to product A[0...i-1]
b[i]: equal to product A[i+1...n)
2) so for each A[i],its product is f[i]*b[i]
eg:[1,2,3,4]
f  [1, 1, 2,6]
b  [24,12,4,1]
r  [24,12,8,6]
time O(n)  space O(n)
*/
vector<int> productExceptSelf(vector<int>& nums) {
vector<int>result;
int n = nums.size();
if(n < 1)return result;
vector<int>f(n,1),b(n,1);
for(int i = 1;i < n;i++){
f[i] = f[i-1]*nums[i-1];
}
for(int i = n - 2;i >= 0;i--){
b[i] = b[i+1]*nums[i+1];
}

for(int i = 0;i < n;i++){
result.push_back(f[i]*b[i]);
}
return result;
}
};
/*algorithm2: optimize the space about f and b
space O(1)
*/
vector<int> productExceptSelf(vector<int>& nums) {
int n = nums.size();
vector<int>result(n);
result[0] = 1;
for(int i = 1;i < n;i++){
result[i] = result[i-1]*nums[i-1];
}
int right = 1;
for(int i = n - 2;i >= 0;i--){
right = right*nums[i+1];
result[i] = right*result[i];
}
return result;
}