题目描述：
给定一个长度为n的整数数组Array【】，输出一个等长的数组result【】，这个输出数组，对应位置i是除了Array【i】之外，其他的所有元素的乘积
例如：
given [1,2,3,4], return [24,12,8,6].
要求：
时间复杂度是o（n）
原文描述：
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.)
思路1：
- 考虑构造两个数组相乘来解决
例如nums=[a1,a2,a3,a4]，构造的数组是：
[1, a1, a1*a2, a1*a2*a3]
[a2*a3*a4, a3*a4, a4, 1]上面的数组相乘，得到[a2*a3*a4, a1*a3*a4, a1*a2*a4, a1*a2*a3]
public class Solution {
public int[] productExceptSelf(int[] nums) {
final int[] result = new int[nums.length];
final int[] left = new int[nums.length];
final int[] right = new int[nums.length];
left[0] = 1;
right[nums.length - 1] = 1;
for (int i = 1; i < nums.length; ++i) {
left[i] = nums[i - 1] * left[i - 1];
}
for (int i = nums.length - 2; i >= 0; --i) {
right[i] = nums[i + 1] * right[i + 1];
}
for (int i = 0; i < nums.length; ++i) {
result[i] = left[i] * right[i];
}
return result;
}
}思路2：（空间复杂度o（1））
- 考虑上面的第二个数组的数据用一个常数代替，然后输出的数组，是不算空间的
代码：
public class Solution {
public int[] productExceptSelf(int[] nums) {
final int[] left = new int[nums.length];
left[0] = 1;
for (int i = 1; i < nums.length; ++i) {
left[i] = nums[i - 1] * left[i - 1];
}
int right = 1;
for (int i = nums.length - 1; i >= 0; --i) {
left[i] *= right;
right *= nums[i];
}
return left;
}
}