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

    题目链接：https://leetcode.com/problems/product-of-array-except-self/

238. Product of Array Except Self

My Submissions

Question

Total Accepted: 36393  Total Submissions: 87262  Difficulty: Medium

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 of nums 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 count as extra space for the purpose of space complexity analysis.)

Subscribe to see which companies asked this question

Show Tags

Show Similar Problems

Have you met this question in a real interview? 

Yes

 

No

Discuss

    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;
	}
}