Product of Array Except Self
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.)
分析
我们以一个4个元素的数组为例,nums=[a1,a2,a3,a4]
,要想在O(n)
的时间内输出结果,比较好的解决方法是提前构造好两个数组:
- [1, a1, a1a2, a1a2*a3]
- [a2a3a4, a3a4, a4, 1]
然后两个数组一一对应相乘,即可得到最终结果[a2
a3a4, a1a3a4, a1a2a4, a1a2*a3]
。
不过,上述方法的空间复杂度为O(n)
,可以进一步优化成常数空间,即用一个整数代替第二个数组。
代码1 O(n)空间
// Product of Array Except Self
// Time Complexity: O(n), Space Complexity: O(n)
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;
}
}
代码2 O(1)空间
// Product of Array Except Self
// Time Complexity: O(n), Space Complexity: O(1)
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;
}
}