238. Product of Array Except Self

• 48.0%

https://leetcode.com/problems/product-of-array-except-self/#/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.)

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剑指offer 52

https://discuss.leetcode.com/topic/20434/o-n-time-and-o-1-space-c-solution-with-explanation

O(n) time and O(1) space C++ solution with explanation

方法一：

First, consider O(n) time and O(n) space solution.

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;
    }
};

方法二：

空间从o（n）降低到o（1）

对从前至后的部分考虑保留前一个值，得到结果。

很巧妙地方法，节省了空间。

We just need to change the two vectors to two integers and note that we should do multiplying operations for two related elements of the results vector in each loop.

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;
    }
};

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https://discuss.leetcode.com/topic/37927/how-from-o-n-to-o-1

How from O(N) to O(1)

Here is the O(N) based C++ implementation

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int len=nums.size();
        vector<int> left(len, 1);
        vector<int> right(len, 1);
        vector<int> result(len, 0);
        for(int i=1; i<len; i++)  left[i]=left[i-1]*nums[i-1];
        for(int i=len-2; i>=0; i--)  right[i]=right[i+1]*nums[i+1];
        for(int i=0; i<len; i++) result[i]=left[i]*right[i];
        return result;
    }
};

How to use O(1) ?

By observing the above code, we can just for every position multiply it to its right position.

Just the idea to think reversly !

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

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https://discuss.leetcode.com/topic/18877/my-c-solution-o-n-time-with-no-extra-space

My C++ solution, O(n) time with no extra space

vector<int> productExceptSelf(vector<int>& nums) {
    int N = nums.size();
    vector<int> res(N,1);
    
    for(int i=0; i<N; i++){
        if (i==0)   res[i] = 1;
        else res[i] = res[i-1]*nums[i-1];
    }
    
    int r_prod = 1;
    for(int i=N-1; i>=0; i--){
        res[i] *= r_prod;
        r_prod *= nums[i];
    }
    
    return res;
}

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https://discuss.leetcode.com/topic/18983/python-solution-accepted-o-n-time-o-1-space

Python solution (Accepted), O(n) time, O(1) space

class Solution:
    # @param {integer[]} nums
    # @return {integer[]}
    def productExceptSelf(self, nums):
        p = 1
        n = len(nums)
        output = []
        for i in range(0,n):
            output.append(p)
            p = p * nums[i]
        p = 1
        for i in range(n-1,-1,-1):
            output[i] = output[i] * p
            p = p * nums[i]
        return output

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https://discuss.leetcode.com/topic/30115/very-easy-two-passes-solution

Very easy two passes solution

// two passes, O(2n)
vector<int> productExceptSelf(vector<int>& nums) {
    int n = nums.size();
    vector<int> ans(n, 1);
    
    for (int i = 1; i < n; ++i) {
        ans[i] = ans[i-1] * nums[i-1];
    }
    
    int m = 1;
    for (int i = n-1; i >= 0; --i) {
        ans[i] *= m;
        m *= nums[i];
    }
    
    return ans;
}