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.)
代码的思路是:给定一个数组a,求数组a的第i位置的除自身之外的元素的乘积,可以将先求出数组a的第1个位置到第i-1个位置的乘积,存到数组b中,在求出第i+1个元素到第n个元素的乘积
public class Solution { public int[] ProductExceptSelf(int[] nums) { int[] result=new int[nums.Length]; int left=1,right=1; result[0]=1; for(int i=1;i<nums.Length;i++) { result[i]= result[i-1]*nums[i-1]; } for(int i=nums.Length-1;i>=0;i--) { result[i]*=right; right*=nums[i]; } return result; } }