Product of Array Except Self

Product of Array Except Self Tota

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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**因为题目要求时间复杂度为O(n^2)且不能用除法，所以不能用先算出sum，再除以num[i]的方法（除数为0，运算不不合法）。

观察图，可知该题目可以用数i左边的数积*数i右边的数积。且可以看做l[0]=1,r[n-1]=1。

最后可知，results[i]=left[i]*right[i].

此算法的复杂度为O(n)**

代码：

class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int i;
int len=nums.size();
vector<int> results,left,right;

if(len < 2) return results;

left[0]=1;
right[len-1]=1;
//左边
for(i=0;i<len;i++){
left[i+1]=left[i]*nums[i];
}
//右边
for(i=len-1;i>=1;i--){
right[i-1]=right[i]*nums[i];
}

for(i=0;i<len;i++){
results[i]=left[i]*right[i];
}
return results;
}
};

可参考：http://blog.csdn.net/yangliuy/article/details/46954779