Product of Array Except Self

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;
}
}