算法设计与应用基础：第一周（3）

238. Product of Array Except Self

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Description Submission Solutions

Total Accepted: 84193
Total Submissions: 176822
Difficulty: Medium
Contributors: Admin

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.)
codes：

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];

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

解题思路：主要在于算法中的两层循环，思路借鉴了博客一位大神对于算数组前n项积的方法。第一层循环：利用哨兵值Fromfirst每次res[i]累乘Fromfirst相当于累乘nums[i]之前的数组累积；第二层循环：res[i]每次累乘Fromlast相当于累乘nums[i]之后的数组累积；两层循环之后可得到正确的res值

res[i]*=fromBegin;
fromBegin*=nums[i];

第一行res[i]累乘fromBegin，第二行fromBegin累乘nums[i]以达到res[i]累乘之前所有的目的