LeetCode 238. Product of Array Except Self

原题链接在这里：https://leetcode.com/problems/product-of-array-except-self/

题目：

Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i].

The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

You must write an algorithm that runs in O(n) time and without using the division operation.

Example 1:

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

Example 2:

Input: nums = [-1,1,0,-3,3]
Output: [0,0,9,0,0] 

Constraints:

• 2 <= nums.length <= 105
• -30 <= nums[i] <= 30
• The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

题解：

看到这道题内心十分憎恨自己，这道题在Amazon电面中遇到，当时没有刷题，没有想到这种方法，若是当时刷了题该多好.

leftArr rightArr的方法早已知晓. follow-up 里说不用额外空间，可以采用res数组间接保留leftArr 和 rightArr 数组. 

两遍iteration, 第一遍更新左侧, res[i] = left; 第二遍更新右侧, res[i] *= right. 注意一边是是等于，一边是乘等于.

Note: 该i--时不要顺手写成i++.

Method 1 Time Complexity: O(n). Space: O(1).

Method 2 Time Complexity: O(n). Space: O(n), res size.

AC Java:

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        //Method 1
        /*
        if(nums == null || nums.length == 0){
            return nums;
        }
        int [] index = new int[2];
        Arrays.fill(index,-1);
        int product = 1;
        
        for(int i = 0; i<nums.length; i++){
            product *= nums[i];
            if(nums[i] == 0 && index[0] == -1){
                index[0] = i;
            }else if(nums[i] == 0 && index[1] == -1){
                index[1] = i;
            }
        }
        if(index[0] == -1){
            for(int i = 0; i<nums.length; i++){
                nums[i] = product/nums[i];
            }
        }else if(index[1] == -1){
            int po = 1;
            for(int i = 0; i<nums.length; i++){
                if(i != index[0]){
                    po*=nums[i];
                }
            }
            Arrays.fill(nums,0);
            nums[index[0]] = po;
        }else{
            Arrays.fill(nums,0);
        }
        
        return nums;
        */
        
        //Method 2
        if(nums == null || nums.length == 0){
            return nums;
        }
        int [] res = new int[nums.length];
        int left = 1;
        int right = 1;
        for(int i = 0; i<nums.length; i++){
            res[i] = left;
            left *=nums[i];
        }
        for(int i = nums.length-1; i>=0; i--){
            res[i] *=right;
            right *=nums[i];
        }
        return res;
    }
}