# Product of Array Except Self

先从左往右扫一遍。res 数组对应的位置是nums对应的数字的左边的乘积，然后再用一个right 变量，然后从数组右往左遍历，同时更新当前遍历数组右边的乘积，当前数组res数组的数是其左边的乘积，左右相乘就是所要结果

Given an array `nums` of *n* integers where *n* > 1,  return an array `output` such that `output[i]` is equal to the product of all the elements of `nums` except `nums[i]`.

**Example:**

```
Input:  [1,2,3,4]
Output: [24,12,8,6]
```

**Note:** Please solve it **without division** and in O(*n*).

**Follow up:**\
Could you solve it with constant space complexity? (The output array **does not** count as extra space for the purpose of space complexity analysis.)

时间复杂度 o(n) 空间复杂度 o(n)

```java
class Solution {
    public int[] productExceptSelf(int[] nums) {
        if(nums == null || nums.length == 0){
            return new int[0];
        }
        
        int[] res = new int[nums.length];
        
        res[0] = 1;
        
        for(int i = 1; i < res.length;i++){
            res[i] = nums[i-1] *res[i-1]; 
        }
        
        int right = 1;
        
        for(int i = res.length-1;i>= 0 ;i--){
            res[i] = right * res[i];
            right = right * nums[i];
        }
        
        return res;
    }
}
```


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