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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; } }
238. Product of Array Except Self
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原文地址:http://www.cnblogs.com/Machelsky/p/5894783.html