【LeetCode】238. Product of Array Except Self

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.)

就是用减法实现除法。

注意零的处理。

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int size = nums.size();
        vector<int> ret(size, );
        long long product = ;
        int countZero = ;
        int ind = -;   // 0-index

        for(int i = ; i < size; i ++)
        {
            if(nums[i] == )
            {
                countZero ++;
                ind = i;
            }
        }

        //special case for 0
        if(countZero == )
        {//no zero
            for(int i = ; i < size; i ++)
                product *= nums[i];
            for(int i = ; i < size; i ++)
                ret[i] = mydivide(product, nums[i]);
        }
        else if(countZero == )
        {//1 zero
            for(int i = ; i < size; i ++)
            {
                if(i != ind)
                    product *= nums[i];
            }
            ret[ind] = product; //others are 0s
        }
        else
        {//2 or more zeros
            ;   //all 0s
        }

        return ret;
    }
    int mydivide(long long product, int divisor)
    {// guaranteed that divisor is not 0
        int sign = ;
        if((product < ) ^ (divisor < ))
            sign = -;
        if(product < )
            product = -product;
        if(divisor < )
            divisor = -divisor;
        //to here, product and divisor are positive
        int ret = ;
        while(true)
        {
            int part = ;   //part quotient
            int num = divisor;
            while(product > num)
            {
                num <<= ;
                part <<= ;
            }
            if(product == num)
            {
                ret += part;
                return sign * ret;
            }
            else
            {
                num >>= ;
                part >>= ;
                ret += part;
                product -= num;
            }
        }
    }
};