Product of Array Exclude Itself详解编程语言

Given an integers array A. 
 
Define B[i] = A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1], calculate B WITHOUT divide operation. 
 
Example 
For A=[1, 2, 3], return [6, 3, 2].

class Solution { 
public: 
    /** 
     * @param A: Given an integers array A 
     * @return: A long long array B and B[i]= A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1] 
     */ 
    vector<long long> productExcludeItself(vector<int> &nums) { 
        const int nums_size = nums.size(); 
        vector<long long> result(nums_size, 1); 
        if (nums.empty() || nums_size == 1) { 
            return result; 
        } 
 
        vector<long long> left(nums_size, 1); 
        vector<long long> right(nums_size, 1); 
        for (int i = 1; i != nums_size; ++i) { 
            left[i] = left[i - 1] * nums[i - 1]; 
            right[nums_size - i - 1] = right[nums_size - i] * nums[nums_size - i]; 
        } 
        for (int i = 0; i != nums_size; ++i) { 
            result[i] = left[i] * right[i]; 
        } 
 
        return result; 
    } 
};

public class Solution { 
 
    public int[] productExceptSelf(int[] A) { 
         
        int[] left = new int[A.length]; 
        int[] right = new int[A.length]; 
        int[] result = new int[A.length]; 
         
        for (int i = 0; i < A.length; i++) { 
            left[i] = i == 0 ? A[i] : left[i - 1] * A[i]; 
        } 
 
        for (int i = A.length - 1; i >= 0; i--) { 
            right[i] = (i == A.length - 1) ? A[i] : right[i + 1] * A[i]; 
        } 
         
        for (int i = 0; i < A.length; i++) { 
            if (i == 0) { 
                result[i] = right[1]; 
            } else if (i == A.length - 1) { 
                result[i] = left[i - 1]; 
            } else { 
                result[i] = left[i - 1] * right[i + 1]; 
            } 
        } 
        return result; 
    } 
}

class Solution { 
public: 
    /** 
     * @param A: Given an integers array A 
     * @return: A long long array B and B[i]= A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1] 
     */ 
    vector<long long> productExcludeItself(vector<int> &nums) { 
        const int nums_size = nums.size(); 
        vector<long long> result(nums_size, 1); 
 
        // solve the left part first 
        for (int i = 1; i < nums_size; ++i) { 
            result[i] = result[i - 1] * nums[i - 1]; 
        } 
 
        // solve the right part 
        long long temp = 1; 
        for (int i = nums_size - 1; i >= 0; --i) { 
            result[i] *= temp; 
            temp *= nums[i]; 
        } 
 
        return result; 
    } 
};