lintcode：Product of Array Exclude Itself

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].

1.提交1

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) {
        // write your code here

        vector<long long> res;
        for(int i=0;ilong long product=1;
            for(int j=0;jif(j!=i){
                    product*=nums[j];
                }
            }
            res.push_back(product);
        }
        return res;
    }
};

AC时间是62ms.时间复杂度O（n^2）

2.提交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) {
        // write your code here
        int n=nums.size();

        long long left[n];
        long long right[n];

        left[0]=1;
        for(int i=1;i1]*nums[i-1];
        }

        right[n-1]=1;
        /*for(int i=0;i
        for(int i=n-2;i>=0;i--){
            right[i]=right[i+1]*nums[i+1];
        }

        vector<long long> res;
        for(int i=0;ireturn res;

    }
};

AC时间12ms。空间复杂度O(N),时间复杂度O（n）

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