Solve it without division and in O(n).
For example, given [1,2,3,4], return [24,12,8,6].
我们创建一个长度和原数组num[]长度一样的数组result[],先从左边开始,让result[0] = 1; result[1] = 1 * num[0], result[2] = num[0] * num[1], result[n] = num[0] * num[1] ...num[n-1]。然后从右边开始,让result[n] = result[n] * 1; result[n - 1] = num[n] * result[n - 1]; result[0] = num[n] * num[n - 1] .... num[1] * result[0]。这样遍历两遍数组就可以得到结果,代码如下:
public class Solution { public int[] productExceptSelf(int[] nums) { int[] result = new int[nums.length]; int left = 1; result[0] = 1; for(int i = 1; i < nums.length; i++) { result[i] = left * nums[i - 1]; left *= nums[i - 1]; } int right = 1; for(int j = nums.length - 1; j >= 0; j--) { result[j] *= right; right *= nums[j]; } return result; } }