LeetCode OJ 238. Product of Array Except Self 解题报告

238. Product of Array Except Self

My SubmissionsQuestionTotal Accepted:36393Total Submissions:87262Difficulty:Medium

Given an array ofnintegers wheren> 1,nums, return an arrayoutputsuch thatoutput[i]is equal to the product of all the elements ofnumsexceptnums[i].

Solve itwithout pisionand 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 arraydoes notcount as extra space for the purpose of space complexity analysis.)

Subscribeto see which companies asked this question

Show Tags

Show Similar Problems

Have you met this question in a real interview?

Yes

No

给出一个数组，要求计算一个新数组，数组里所有的元素都是除了自己以外的元素乘积。并且要求不许用除法。

《编程之美》上的一道原题。创建两个辅助数组，一个保存所有左边元素乘积的结果。一个保存所有右边元素乘积的结果。借助这两个数组，一次遍历就可以得到结果。

我的AC代码

public class ProductofArrayExceptSelf {

	public static void main(String[] args) {
		int[] a = { 1, 2, 3, 4 };
		System.out.print(Arrays.toString((productExceptSelf(a))));
	}

	public static int[] productExceptSelf(int[] nums) {
		int len = nums.length;
		int[] r = new int[len];

		int[] left = new int[len];
		int[] right = new int[len];
		left[0] = nums[0];
		for (int i = 1; i < len; i++) {
			left[i] = left[i - 1] * nums[i];
		}
		right[len - 1] = nums[len - 1];
		for (int i = len - 2; i >= 0; i--) {
			right[i] = right[i + 1] * nums[i];
		}

		r[0] = right[1];
		r[len - 1] = left[len - 2];
		for (int i = 1; i < len - 1; i++) {
			r[i] = left[i - 1] * right[i + 1];
		}
		return r;
	}
}