Product of Array Except Self - LeetCode

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.)

思路：构建结果数组res，然后从头到尾扫描一遍nums数组。

res[i]所存的值为nums[0]到nums[i - 1]的乘积，res[0] = 1。

最后从尾到头再扫描一遍nums数组，res[i]这次再乘上nums[i + 1]到nums[n - 1]的值。

所乘上的值可以只由一个变量来计算完成。这里用一个int变量，命名为right，初始为1。

则res[i]乘以right就是最后结果值。要注意的是，每次乘完之后，right要更新为right * nums[i]。

class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums){
int n = nums.size();
vector<int> res(n, 1);
for (int i = 0; i < n; i++)
res[i] = (i == 0 ? 1 : res[i - 1] * nums[i - 1]);
int right = 1;
for (int i = n - 1; i >= 0; i--)
{
res[i] *= right;
right *= nums[i];
}
return res;
}
};