【Python】LeetCode 238. Product of Array Except Self

kaka 發布于2019-07-31 12:22 / 1706人閱讀

摘要：題目描述題目解析簡單來說就是對于數組中每一項，求其他項之積。算一遍全部元素的積再分別除以每一項要仔細考慮元素為零的情況。沒有零直接除下去。一個零零的位置對應值為其他元素之積，其他位置為零。兩個以上的零全部都是零。

題目描述

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].

題目解析

簡單來說就是對于數組中每一項，求其他項之積。

解題思路 對于每一項硬算其他項之積

恭喜你，你超時了。

算一遍全部元素的積再分別除以每一項

要仔細考慮元素為零的情況。

沒有零

直接除下去。

一個零

零的位置對應值為其他元素之積，其他位置為零。

兩個以上的零

全部都是零。

AC代碼

class Solution(object):
    def productExceptSelf(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        try:
            from functools import reduce
        finally:
            pass
        res = []
        zeros = nums.count(0)
        if zeros == 0:
            product = reduce(lambda x, y: x * y, nums)
            res = [product // x for x in nums]
        elif zeros == 1:
            now = nums[::]
            pos = now.index(0)
            del now[pos]
            product = reduce(lambda x, y: x * y, now)
            res = [0 if x != pos else product for x in range(len(nums))]
        else:
            res = [0] * len(nums)
        return res

總結

遇事多思考，輕易不要循環。

GPU云服務器云服務器 Except LeetCode Product LeetCode_String

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