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.)
简单方法是使用两个数组a,b，其中一个存储nums[0..i - 1], 另一个数组存nums[i + 1..n-1], 于是结果为result[i] = a[i] * b[n - i - 1]，即
a存储i前面的积，b存储i后面元素的积
vector<int> productExceptSelf(vector<int>& nums) {
if (nums.empty()) // 空
return vector<int>();
if (nums.size() == 1) { //只有一个元素，返回{0}
return vector<int>({0});
}
auto n = nums.size();
vector<int> a(n, 1), b(n, 1);
for (auto i = 1; i < n; ++i) {
a[i] = a[i - 1] * nums[i - 1];
b[i] = b[i - 1] * nums[n - i];
}
for (auto i = 0; i < n; ++i) {
a[i] *= b[n - i - 1];
}
return a;
}以上方法O(n)时间，除了结果空间，还需要额外空间O(n)
题目要求O(1)空间。
我们发现a数组得到i前面的积后，只需要累乘后面的积即可。
a[n- 1]不需要乘a[n - 2] = a[n - 2] * nums[n - 1]a[n - 3] = a[n - 3] * nums[n - 1] * nums[n - 2]- ...
a[0] = a[0] * nums[n - 1] * nums[n - 2] * ... * nums[1]
使用一个变量product，表示从后往前的累乘，则a[i] = a[i] * product
vector<int> productExceptSelf(vector<int> &nums) {
if (nums.empty()) // 空
return vector<int>();
if (nums.size() == 1) { //只有一个元素，返回{0}
return vector<int>({0});
}
int n = nums.size();
vector<int> result(n, 1);
for (int i = 1; i < n; ++i)
result[i] = result[i - 1] * nums[i - 1];
int product = 1;
for (int i = n - 2; i >= 0; --i) {
product *= nums[i + 1];
result[i] *= product;
}
return result;
}此时空间为O(1)