# 238.Product-of-Array-Except-Self

## 238. Product of Array Except Self

## 题目地址

<https://leetcode.com/problems/product-of-array-except-self/>

## 题目描述

```
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.)
```

## 代码

### Approach 1: Left + Right Array

```java
class Solution {
    public int[] productExceptSelf(int[] nums) {
    int length = nums.length;
    int[] L = new int[length];
    int[] R = new int[length];

    int[] answer = new int[length];

    L[0] = 1;
    for (int i = 1; i < length; i++) {
      L[i] = nums[i - 1] * L[i - 1];
    }

    R[length - 1] = 1;
    for (int i = lenght - 2; i >= 0; i--) {
      R[i] = nums[i + 1] * R[i + 1];
    }

    for (int i = 0; i < length; i++) {
      answer[i] = L[i] * R[i];
    }

    return answer;
  }

}
```

### Approach 2: O(1) space approach

```java
class Solution {
  public int[] productExceptSelf(int[] nums) {
    int length = nums.length;
    int[] answer = new int[length];

    answer[0] = 1;
    for (int i = 1; i < length; i++) {
      answer[i] = nums[i - 1] * answer[i - 1];
    }

    int R = 1;
    for (int i = length - 1; i >= 0; i--) {
      answer[i] = answer[i] * R;
      R *= nums[i];
    }

    return answer;
  }
}
```


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