LintCode 1310: Product of Array (数组好题)

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分类专栏： algorithm-design 文章标签： LintCode

于 2019-01-11 13:46:52 首次发布

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本文探讨了在不使用除法的情况下，计算数组中每个元素的乘积，但排除当前元素的方法。通过两次遍历数组，一次从左到右记录前缀乘积，一次从右到左记录后缀乘积，最终得到所有元素的除自身外乘积。

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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].

Example
For example, given [1,2,3,4], return [24,12,8,6].

Challenge
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.)

Notice
Solve it without division and in O(n).

解法1：
思路：用两个array: forward 和 backward。

代码如下：

class Solution {
public:
    /**
     * @param nums: an array of integers
     * @return: the product of all the elements of nums except nums[i].
     */
    vector<int> productExceptSelf(vector<int> &nums) {
        
        int len = nums.size();
        
        if (len == 0) return vector<int>();
        if (len == 1) return nums;
        
        vector<int> result(len, 0);
        vector<int> forward(len, 0);
        vector<int> backward(len, 0);
        
        forward[0] = nums[0];
        for (int i = 1; i < len; ++i) {
            forward[i] = forward[i - 1] * nums[i];
        }

        backward[len - 1] = nums[len - 1];
        for (int i = len - 2; i >= 0; --i) {
            backward[i] = backward[i + 1] * nums[i];
        }
        
        for (int i = 1; i < len - 1; ++i) {
            result[i] = forward[i - 1] * backward[i + 1];
        }
        result[0] = backward[1];
        result[len - 1] = forward[len - 2];
        
        return result;
    }
};

解法2：一个array，直接基于return array操作。反向数组用一个变量代替即可。
代码如下：

class Solution {
public:
    /**
     * @param nums: an array of integers
     * @return: the product of all the elements of nums except nums[i].
     */
    vector<int> productExceptSelf(vector<int> &nums) {
        
        int len = nums.size();
        
        if (len == 0) return vector<int>();
        if (len == 1) return nums;
        
        vector<int> result(len, 1);
        for (int i = 1; i < len; ++i) { 
             //note result[i] only records the forward product of i - 1
            result[i] = result[i - 1] * nums[i - 1];
        }

        int backwardProduct = 1;
        for (int i = len - 1; i >= 0; --i) {
            result[i] = result[i] * backwardProduct;
            backwardProduct *= nums[i];
        }
         
        return result;
    }
};

二刷:

class Solution {
public:
    /**
     * @param nums: an array of integers
     * @return: the product of all the elements of nums except nums[i].
     */
    vector<int> productExceptSelf(vector<int> &nums) {
        int n = nums.size();
        vector<int> res(n, 1);
        res[0] = nums[0];
        for (int i = 1; i < n - 1; i++) {
            res[i] = res[i - 1] * nums[i];
        }
        int backwardProd = 1;
        for (int i = n - 1; i >= 1; i--) {
            res[i] = res[i - 1] * backwardProd;
            backwardProd *= nums[i];
        }
        res[0] = backwardProd;
        return res;
    }
};