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=,right=;
result[]=;
for(int i=;i<nums.Length;i++)
{
result[i]= result[i-]*nums[i-];
} for(int i=nums.Length-;i>=;i--)
{
result[i]*=right;
right*=nums[i];
} return result;
}
}