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product_of_array_except_self.cpp
54 lines (47 loc) · 1.58 KB
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product_of_array_except_self.cpp
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/**
* Product of array except itself
* Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].
Solve it without division and in O(n).
For example, given [1,2,3,4], return [24,12,8,6].
Follow up:
Could you solve it with constant space complexity? (Note: The output array does not count as extra space for the purpose of space complexity analysis.)
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*/
#include <iostream>
#include <vector>
using namespace std;
namespace product_of_array_except_itself {
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
if (nums.empty())
return vector<int>();
vector<int> ans(nums.size(), 1);
for (int i = nums.size()-1; i >= 0; i--) {
if (i == nums.size()-1)
ans[i] = nums[i];
else
ans[i] = ans[i+1]*nums[i];
}
for (int i=1; i<nums.size(); i++) {
nums[i] = nums[i-1] * nums[i];
}
for (int i = 0; i < ans.size(); i++) {
if (i == 0)
ans[i] = ans[i+1];
else if (i == ans.size()-1)
ans[i] = nums[i-1];
else
ans[i] = nums[i-1]*ans[i+1];
}
return ans;
}
};
}
int main_product_of_array_except_itself() {
int data[] = {1,2,3,4};
vector<int> nums(data, data+4);
vector<int> ans = (new product_of_array_except_itself::Solution)->productExceptSelf(nums);
for (auto num : ans)
cout << num << " ";
}