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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, a1*a2, a1*a2*a3]
  • [a2*a3*a4, a3*a4, a4, 1]

然后两个数组一一对应相乘,即可得到最终结果 [a2*a3*a4, a1*a3*a4, a1*a2*a4, a1*a2*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;
}
}