【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, 0);
        long long product = 1;
        int countZero = 0;
        int ind = -1;   // 0-index
        
        for(int i = 0; i < size; i ++)
        {
            if(nums[i] == 0)
            {
                countZero ++;
                ind = i;
            }
        }
        
        //special case for 0
        if(countZero == 0)
        {//no zero
            for(int i = 0; i < size; i ++)
                product *= nums[i];
            for(int i = 0; i < size; i ++)
                ret[i] = mydivide(product, nums[i]);
        }
        else if(countZero == 1)
        {//1 zero
            for(int i = 0; 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 = 1;
        if((product < 0) ^ (divisor < 0))
            sign = -1;
        if(product < 0)
            product = -product;
        if(divisor < 0)
            divisor = -divisor;
        //to here, product and divisor are positive
        int ret = 0;
        while(true)
        {
            int part = 1;   //part quotient
            int num = divisor;
            while(product > num)
            {
                num <<= 1;
                part <<= 1;
            }
            if(product == num)
            {
                ret += part;
                return sign * ret;
            }
            else
            {
                num >>= 1;
                part >>= 1;
                ret += part;
                product -= num;
            }
        }
    }
};