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  • 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.)

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        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;
    	}
    }
    


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  • 原文地址:https://www.cnblogs.com/llguanli/p/8946640.html
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