Product of Array Except Self

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问题描述

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.)

解决思路

1. 使用辅助空间

为方便理解，使用两个与输入数组同等大小的辅助数组，

首先，从前到后扫描数组，一个数组[i]存储的是除了当前元素外的所有前面元素的乘积；

然后，从后到前扫描数组，另一个数组[i]存储的是除了当前元素外的所有后面元素的乘积。

最后将这两个数组的对应位置的元素相乘即可。

优化后可以省略一个数组和一遍扫描。

程序

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        if (nums == null) {
            return null;
        }
        int len = nums.length;
        int[] res = new int[len];
        int[] tmp = new int[len];
        tmp[0] = 1;
        res[0] = 1;
        
        for (int i = 1; i < len; i++) {
            tmp[i] = tmp[i - 1] * nums[i - 1];
            res[i] = tmp[i];
        }
        tmp[len - 1] = 1;
        for (int i = len - 2; i >= 0; i--) {
            tmp[i] = tmp[i + 1] * nums[i + 1];
            res[i] *= tmp[i];
        }
        
        return res;
    }
}

Product of Array Except Self

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原文地址：http://www.cnblogs.com/harrygogo/p/4650485.html