LeetCode-238. Product of Array Except Self

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本文介绍了一种不使用除法且时间复杂度为O(n)的方法来计算除自身外的所有元素乘积。该方法通过两次遍历数组实现，分别从前向后和从后向前计算累积乘积。

Description

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.)

Solution 1(C++)

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n=nums.size();
        vector<int> fromBegin(n);
        fromBegin[0]=1;
        vector<int> fromLast(n);
        fromLast[0]=1;

        for(int i=1;i<n;i++){
            fromBegin[i]=fromBegin[i-1]*nums[i-1];
            fromLast[i]=fromLast[i-1]*nums[n-i];
        }

        vector<int> res(n);
        for(int i=0;i<n;i++){
            res[i]=fromBegin[i]*fromLast[n-1-i];
        }
        return res;
    }
};

Solution 2(C++)

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n=nums.size();
        int fromBegin=1;
        int fromLast=1;
        vector<int> res(n,1);

        for(int i=0;i<n;i++){
            res[i]*=fromBegin;
            fromBegin*=nums[i];
            res[n-1-i]*=fromLast;
            fromLast*=nums[n-1-i];
        }
        return res;
    }
};