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
;
long long right
;

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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