[Leetcode] Product of array Except Self in addition to the product of an array

Given an array of n integers where n > 1, nums and return an array output such that's equal to output[i] th E 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 solve it with constant space complexity? (note:the output array does not count as extra space for the purpose of space complexity analysis.)

Given an array of us, let's return a new array, the number of each position is the product of the number of other positions, and limit the time complexity O (n), and do not let us use division. If you use division, then this problem should be easy, because you can go through the array to find out the product of all numbers, and then divided by the number of the corresponding position. But this problem forbids us to use division, then we can only go the other way. We can walk through the array first, and the product of all the numbers before each position. Then again, the last position on the number is the product of all the previous numbers, is in line with the title, but all the previous numbers need to continue to multiply. We're going to scan back and forth, multiplying the number in each position by the product of all the numbers in the back, and for the last position, multiply by 1, because there are no numbers behind it. See the code below:

classSolution { Public: Vector<int> productexceptself (vector<int>&nums) {Vector<int> Res (nums.size (),1);  for(inti =1; I < nums.size (); ++i) {res[i]= Res[i-1] * Nums[i-1]; }        intright =1;  for(inti = nums.size ()-1; I >=0; --i) {res[i]*=Right ; Right*=Nums[i]; }        returnRes; }};

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[Leetcode] Product of array Except Self in addition to the product of an array