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;i<nums.size();i++){            long long product=1;            for(int j=0;j<nums.size();j++){                if(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;i<n;i++){            left[i]=left[i-1]*nums[i-1];        }        right[n-1]=1;        /*for(int i=0;i<n-1;i++){            right[i]=right[i+1]*nums[i+1];        }*/        for(int i=n-2;i>=0;i--){            right[i]=right[i+1]*nums[i+1];        }        vector<long long> res;        for(int i=0;i<n;i++){            res.push_back(left[i]*right[i]);        }        return res;    }};

AC时间12ms。空间复杂度O(N),时间复杂度O（n）

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