【leetcod】Product of Array Except Self -C++

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.)

大意：给一个数组a，然后返回一个数组b，b数组中的数是对应的是a数组中除了自己以外剩下所有数的乘积。不能使用除法。

假设数组为：

a1，a2，a3，a4

可以创建两个有规律数组

1，a1，a1a2,，a1a2a3

a2a3a4，a3a4，a4,1

然后对应相乘就可以了。

还可以只创建一个数组 1，a1，a1a2,，a1a2a3，此时最后一项已经对了，只差前面的了，然后逆序遍历，用一个临时变量来存储要乘的数据。

class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
if(nums.size() <= 1)
{
vector<int> ivec(nums.begin(),nums.end());
return ivec;
}
vector<int> ivec(nums.size(),1);
for(int i = 1;i < nums.size();++i)
{
ivec[i] = ivec[i-1] * nums[i-1];
}
int numtmp = 1;
for(int i = nums.size() - 1; i >= 0;--i)
{
ivec[i] = ivec[i] * numtmp;
numtmp *= nums[i];
}
return ivec;

}
};