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.)

 

思路:类似于dp的想法,先用left array存数字左边的所有乘积。然后再从右至左乘上右边的乘积。

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int[] left=new int[nums.length];
        left[0]=1;
        for(int i=1;i<nums.length;i++)
        {
            left[i]=left[i-1]*nums[i-1];
        }
        int right=1;
        for(int i=nums.length-1;i>=0;i--)
        {
            left[i]*=right;
            right*=nums[i];
        }
        return left;
    }
}

 

posted on 2016-09-22 05:10  Machelsky  阅读(107)  评论(0)    收藏  举报