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LeetCode OJ 238. Product of Array Except Self 解题报告

product of array except self

    题目链接:https://leetcode.com/problems/product-of-array-except-self/

238. Product of Array Except Self

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Total Accepted: 36393  Total Submissions: 87262  Difficulty: Medium

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.)

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    给出一个数组,要求计算一个新数组,数组里所有的元素都是除了自己以外的元素乘积。并且要求不许用除法。

    《编程之美》上的一道原题。创建两个辅助数组,一个保存所有左边元素乘积的结果。一个保存所有右边元素乘积的结果。借助这两个数组,一次遍历就可以得到结果。

    我的AC代码

  1. public class ProductofArrayExceptSelf {
  2. public static void main(String[] args) {
  3. int[] a = { 1, 2, 3, 4 };
  4. System.out.print(Arrays.toString((productExceptSelf(a))));
  5. }
  6. public static int[] productExceptSelf(int[] nums) {
  7. int len = nums.length;
  8. int[] r = new int[len];
  9. int[] left = new int[len];
  10. int[] right = new int[len];
  11. left[0] = nums[0];
  12. for (int i = 1; i < len; i++) {
  13. left[i] = left[i - 1] * nums[i];
  14. }
  15. right[len - 1] = nums[len - 1];
  16. for (int i = len - 2; i >= 0; i--) {
  17. right[i] = right[i + 1] * nums[i];
  18. }
  19. r[0] = right[1];
  20. r[len - 1] = left[len - 2];
  21. for (int i = 1; i < len - 1; i++) {
  22. r[i] = left[i - 1] * right[i + 1];
  23. }
  24. return r;
  25. }
  26. }


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