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 ofnums 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

題目大意：某一整型陣列，對於其中的每個元素，計算陣列中除了該元素之外所有元素的積，用陣列返回結果。

題目思路：

解法一：先對於每一個元素，計算這個元素左邊所有元素的積，再用該乘積乘上該元素右邊所有元素的乘積。重點是採取累乘的作法，一共只遍歷陣列兩遍，避免O(n*n)的時間複雜度。程式碼如下：（2ms，beats 37.78%）

public int[] productExceptSelf(int[] nums) {
        int len = nums.length,rightProduct=1;
		int[] output = new int[len];
		output[0] = 1;
		for (int i = 1; i < len; i++) {
			output[i] = output[i - 1] * nums[i - 1];
		}

		for (int i = len - 1; i >= 0; i--) {
			output[i] *= rightProduct;
			rightProduct *= nums[i];
		}
		return output;
    }
 

public int[] productExceptSelf(int[] nums) {
int len = nums.length, product = 1;
int[] output = new int[len];
for (int num : nums)
product *= num;
for (int i = 0; i < len; i++) {
if (nums[i] == 0) {
int k = 1;
for (int j = 0; j < i && k != 0; j++)
k *= nums[j];
for (int j = i + 1; j < len && k != 0; j++) {
k *= nums[j];
output[j] = 0;
}
output[i] = k;
return output;
}
output[i] = product / nums[i];
}
return output;
}