LeetCode 238 Product of Array Except Self 前缀积&后缀积

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.

Solution

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;
    }
};

标签：pre,suf,Product,nums,int,Self,Except,vector,ans
来源： https://www.cnblogs.com/xinyu04/p/16535724.html