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.)

（類似）dp

每一個元素的對應值 等於 它左邊所有元素的乘積 * 它右邊所有元素的乘積

用res[i]表示每一個元素對應值。元素nums[i]左邊所有元素的乘積是left[i-1] * nums[i-1]，第一個元素左邊沒有元素，即乘積為1，left[0] = 1，遍歷一遍得到元素左邊所有元素的乘積。同樣，元素nums[i]右邊所有元素的乘積是right[i+1] * nums[i+1]，最後一個元素右邊沒有元素，即乘積為1，right[nums.length-1] = 1。最後再把left[i] * right[i] 得到res[i]

time: O(n), space: O(n)

class Solution {
    public int[] productExceptSelf(int[] nums) {
        if(nums == null || nums.length == 0) return nums;
        int[] left = new int[nums.length];
        int[] right = new int[nums.length];
        int[] res = new int[nums.length];
        
        left[
 
0] = 1;
        for(int i = 1; i < nums.length; i++) {
            left[i] = left[i-1] * nums[i-1];
        }
        
        right[nums.length - 1] = 1;
        for(int i = nums.length - 2; i >= 0; i--) {
            right[i] = right[i+1] * nums[i+1];
        }
        
        for(int i = 0; i < nums.length; i++) {
            res[i] = left[i] * right[i];
        }
        return res;
    }
}

優化：只用一個res array, space complexity O(1)

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