Given an integer array nums
, return an array answer such that answer[i]
is equal to the product of all the elements of nums except nums[i]
.
The product of any prefix or suffix of nums
is guaranteed to fit in a 32-bit integer.
You must write an algorithm that runs in \(O(n)\) time and without using the division operation.
当前位置的答案就是这个位置 \(pre\times suf\),我们可以迭代更新 \(pre,suf\) 不用额外的储存空间
class Solution { public: vector<int> productExceptSelf(vector<int>& nums) { vector<int> ans(nums.size(),1); int n = nums.size(); int pre=1, suf=1; for(int i=0;i<n;i++){ ans[i]*=pre; pre*=nums[i]; } for(int i=n-1;i>=0;i--){ ans[i]*=suf;suf*=nums[i]; } return ans; } };