C#解leetcode 238. Product of Array Except Self

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;
    }
}

C#解leetcode 238. Product of Array Except Self

原文：http://www.cnblogs.com/xiaohua92/p/5293632.html