LeetCode_238_Product of Array Except Self

238. Product of Array Except Self

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.)

 

题目分析：

    给定一个num数组，n>1，输出一个output数组，且output[i]等于除num[i]之外所有元素的乘积，给出一个满足一下条件的solution:

1、 不使用除法

2、 时间复杂度O(n)

3、 不使用额外的储存空间（输出数组不算）

 

解

假定

s1[0] = nums[0];

s2[n] = nums[n];

构建以下数组

s1[i] = nums[0]* nums[1] * nums[i];

s2[i] = nums[n]* nums[n-1] *...* nums[i];

则可知

output[i] =s1[i-1] * s2[i+1]

其中：

s1[i- 1] = nums[0] * nums[1]*...nums[i-1] 

s2[i+1] = nums[i+1]* nums[i+2]* ... nums[n]

 

Solution1

vector<int> productExceptSelf(vector<int>& nums) {
        int n = nums.size()-1;
        vector<int> vS1(n+1),vS2(n+1);
        vector<int> vRst(n+1);
        int result_s = 1,result_e = 1;
        
        for(int i = 1; i<= n; i++)
            result_s *= nums[i];
        vRst[0] = result_s;
        
        for(int i = 0; i< n; i++)
            result_e *= nums[i];
        vRst[n] = result_e;
        
        vS1[0] = nums[0];
        vS2[n] = nums[n];
        
        for(int i = 1; i<= n; i++)
        {
            vS1[i] = vS1[i-1] * nums[i];　　//由于vS1[0]已知,从vS1[1]开始计算
            vS2[n-i] = vS2[n-i+1] * nums[n-i];　　//由于vS2[n]已知,从vS2[n-1]开始计算
        }
        
        for(int i =1; i< n; i++)
        {
            vRst[i] = vS1[i-1] * vS2[i+1];
        }
        return vRst;
    }

分析两个for循环可知：

1、在第i次循环时，vS1[i-1]是已知的，且vRst[i]的值不会对vS2[i+1]造成影响。

2、所以可将vS1[i-1]用一个int类型变量保存，vS2[i+1]的值则保存为vRst[i+1],以满足题目中不开辟额外空间的要求。

给出以下

Solution2

 vector<int> productExceptSelf(vector<int>& nums) {
        int n = nums.size()-1;
        vector<int> vRst(n+1);
        int result_s = 1;
        
        int s1 = nums[0];
        vRst[n] = nums[n];
        
        for(int i= 1; i<=n; i++)
            vRst[n-i] = vRst[n-i+1] * nums[n-i];
        vRst[0] = vRst[1];
        
        for(int i =1; i<n;i++)
        {
            vRst[i] = s1 *vRst[i+1];
            s1 *= nums[i];
        }
    
        vRst[n] = s1;
        return vRst;
    }

最后是LeetCode Discuss中大犇 给出的答案，比Solution2更快（虽然3个solution Tn = O(n)）

Solution3

    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;
    }