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

  1. // Product of Array Except Self
  2. // Time Complexity: O(n), Space Complexity: O(n)
  3. public class Solution {
  4. public int[] productExceptSelf(int[] nums) {
  5. final int[] result = new int[nums.length];
  6. final int[] left = new int[nums.length];
  7. final int[] right = new int[nums.length];
  8. left[0] = 1;
  9. right[nums.length - 1] = 1;
  10. for (int i = 1; i < nums.length; ++i) {
  11. left[i] = nums[i - 1] * left[i - 1];
  12. }
  13. for (int i = nums.length - 2; i >= 0; --i) {
  14. right[i] = nums[i + 1] * right[i + 1];
  15. }
  16. for (int i = 0; i < nums.length; ++i) {
  17. result[i] = left[i] * right[i];
  18. }
  19. return result;
  20. }
  21. }

代码2 O(1)空间

  1. // Product of Array Except Self
  2. // Time Complexity: O(n), Space Complexity: O(1)
  3. public class Solution {
  4. public int[] productExceptSelf(int[] nums) {
  5. final int[] left = new int[nums.length];
  6. left[0] = 1;
  7. for (int i = 1; i < nums.length; ++i) {
  8. left[i] = nums[i - 1] * left[i - 1];
  9. }
  10. int right = 1;
  11. for (int i = nums.length - 1; i >= 0; --i) {
  12. left[i] *= right;
  13. right *= nums[i];
  14. }
  15. return left;
  16. }
  17. }

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/java/linear-list/array/product-of-array-except-self.html