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
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。数组里全部的元素都是除了自己以外的元素乘积。而且要求不许用除法。

    《编程之美》上的一道原题。创建两个辅助数组。一个保存全部左边元素乘积的结果。一个保存全部右边元素乘积的结果。借助这两个数组,一次遍历就能够得到结果。

    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;
	}
}


posted @ 2018-04-25 19:08  llguanli  阅读(148)  评论(0编辑  收藏  举报