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.)
两次循环,第一次循环得出前面所有元素之积,第二次乘以后面所有元素之积。
class Solution: # @param {integer[]} nums # @return {integer[]} def productExceptSelf(self, nums): result = [0 for x in range(len(nums))] before = after = 1 for i in range(len(nums)): result[i] = before before *= nums[i] for i in range(len(nums)-1,-1,-1): result[i] *= after after *= nums[i] return result
Leetcode 238 Product of Array Except Self
原文:http://www.cnblogs.com/lilixu/p/4657032.html