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product-of-array-except-self

admin 11月 12, 2020 0

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.)

Solution 1.

public class  {
    public int[] productExceptSelf(int[] nums) {
        int products = 1;  
        int zeroNum = 0;　 //０的数目
        int[] output = new int[nums.length];　
        
        for (int num : nums) {
            products *= num;
            if (num == 0)
                zeroNum++;　//统计0的数目
        }
        
        for (int i = 0; i < nums.length; i++) {
        　　//特殊情况，nums[i]=0
            if (nums[i] == 0) {
            　　//除了本身之外还有0，所以except self之后为０
                if (zeroNum > 1) 
                    output[i] = 0;
                //只有self为０，再遍历一遍
                else {
                    output[i] = 1;
                    for (int j = 0; j < nums.length; j++) 
                        if (j != i)
                            output[i] *= nums[j];
                }
            }
            else output[i] = products / nums[i];
        }
        
        return output;
    }
}

Solution 2.

class  {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n = nums.size(), i = 0;
        vector<int> output(n, 1);
        int k = nums[n-1];
        //计算self左边的乘积
        for (i = 1; i < n; ++i)
            output[i] = output[i-1] * nums[i-1];
        //　计算self右边的乘积
        for (i = n - 2; i >=0; --i)
        {
            output[i] *= k;
            k *= nums[i];
        }
        return output;
    }
};