c_cpp 获取一系列数字，并将每个项目替换为数组中所有其他项目的产品

Posted 2021-05-14

#define ARRAY_MAX 30

// forget to consider integer overflow

void product_of_all_other_items_in_array(int A[], int N) {
    int aux1[ARRAY_MAX], aux2[ARRAY_MAX];
    
    aux1[0] = 1;
    for(int i=1; i<N; i++)
        aux1[i] = aux1[i-1] * A[i-1];
    
    aux2[N-1] = 1;
    for(int i=N-2; i>=0; i--)
        aux2[i] = aux2[i+1] * A[i+1]; // gist, should be A[i+1] not A[i-1]
        
    for(int i=0; i<N; i++) 
        A[i] = aux1[i] * aux2[i];
}

// solution 2, use less space
  vector<int> productExceptSelf(vector<int>& nums) {
        vector<int> res(nums.size(), 1);
        for (int i = 1; i < nums.size(); ++i) {
            res[i] = res[i - 1] * nums[i - 1];
        }
        int right = 1;
        for (int i = nums.size() - 1; i >= 0; --i) {
            res[i] *= right;
            right *= nums[i];
        }
        return res;
    }