【LeetCode】238. Product of Array Except Self

最新推荐文章于 2021-10-11 11:07:24 发布

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最新推荐文章于 2021-10-11 11:07:24 发布

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本文介绍了一种高效算法，该算法能在O(n)的时间复杂度内计算出一个整数数组中每个元素对应的除自身以外所有元素的乘积。特别地，文章详细探讨了如何在不使用除法操作的情况下实现这一目标，并考虑了数组中存在零值的特殊情况。

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;
            }
        }
    }
};