Product of Array Exclude Itself

Given an integers array A.

Define B[i] = A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1], calculate B WITHOUT divide operation.

Example

For A = [1, 2, 3], return [6, 3, 2].

分析：

 1 public class Solution {
 2     public int[] productExceptSelf(int[] A) {
 3         int[] left = new int[A.length];
 4         int[] right = new int[A.length];
 5         int[] result = new int[A.length];
 6         
 7         for (int i = 0; i < A.length; i++) {
 8             left[i] = i == 0 ? 1 : left[i - 1] * A[i - 1];
 9             right[A.length - 1 - i] = (i == 0) ? 1 : right[A.length - i] * A[A.length - i];
10         }
11         
12         for (int i = 0; i < A.length; i++) {
13             result[i] = left[i] * right[i];
14         }
15         return result;
16     }
17 }

 

 1 public class Solution {
 2     public int[] productExceptSelf(int[] nums) {
 3         int n = nums.length;
 4         int[] res = new int[n];
 5         res[0] = 1;
 6         for (int i = 1; i < n; i++) {
 7             res[i] = res[i - 1] * nums[i - 1];
 8         }
 9         int right = 1;
10         for (int i = n - 1; i >= 0; i--) {
11             res[i] *= right;
12             right *= nums[i];
13         }
14         return res;
15     }
16 }

 one pass

 1 class Solution {
 2     public int[] productExceptSelf(int[] nums) {
 3         int[] result = new int[nums.length];
 4         Arrays.fill(result, 1);
 5         int left = 1, right = 1;
 6         for (int i = 0, j = nums.length - 1; i < nums.length - 1; i++, j--) {
 7             left *= nums[i];
 8             right *= nums[j];
 9             result[i + 1] *= left;
10             result[j - 1] *= right;
11         }
12         return result;
13     }
14 }

 

posted @ 2016-07-01 11:05  北叶青藤  阅读(366)  评论(0)  编辑  收藏  举报