Leetcode 238 Product of Array Except Self

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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.)

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

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原文地址：http://www.cnblogs.com/lilixu/p/4657032.html