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

Posted 2020-06-17 bruce128

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

238. Product of Array Except Self

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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.)

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    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;
	}
}