python练习（十二）

注意，答案只是代表是他人写的代码，正确，但不一定能通过测试（比如超时），列举出来只是它们拥有着独到之处，虽然大部分确实比我的好

Move Zeroes

题目

Given an array nums, write a function to move all 0’s to the end of it while maintaining the relative order of the non-zero elements.

For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].

Note:
You must do this in-place without making a copy of the array.
Minimize the total number of operations.

思路与解答

class Solution(object):    def moveZeroes(self, nums):        """        :type nums: List[int]        :rtype: void Do not return anything, modify nums in-place instead.        """        l = len(nums)        nums.remove(k)        nums += [0]*(l-len(nums))

remove不是全删掉啊

class Solution(object):    def moveZeroes(self, nums):        """        :type nums: List[int]        :rtype: void Do not return anything, modify nums in-place instead.        """        l = len(nums)        while 0 in nums:            nums.remove(0)        nums += [0]*(l-len(nums))

225ms,15%，绝对不行啊

答案

class Solution(object):    def moveZeroes(self, nums):        """        :type nums: List[int]        :rtype: void Do not return anything, modify nums in-place instead.        """        nums.sort（key = lambda x：1 if x == 0 else 0）

解释

# in-placedef moveZeroes(self, nums):    zero = 0  # records the position of "0"    for i in xrange(len(nums)):        if nums[i] != 0:            nums[i], nums[zero] = nums[zero], nums[i]            zero += 1

Sum of Two Integers

题目

Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.

思路与解答

不允许使用加和减……
怎么做，位操作可以吗？不过不会啊

感谢LHearen，这个链接是关于位操作的一个总结
学习后的代码

        while b:            a,b=a^b,(a&b)<<1              return a

然后测试时输入个负数，炸了
负数是什么鬼东西啊，负数是怎么在python中存储的啊

扩展阅读-取模(mod)与取余(rem)的区别

想搜负数在python中的存储方式，不太好搜，倒是搜到一些其它有意思的东西
负数除法相关-知乎-有些乱，详实
负数除法相关-博客-清晰，不全
取模运算-百度百科-写的不好，我都想把这个链接踢掉
取模(mod)与取余(rem)的区别-鬼知道作者是谁了

上边的扩展阅读和这道题毛线关系都没有，继续做题

这才是对这道题有帮助的文章

#不用额外的变量实现两个数字互换。def swap(num_1, num_2):    num_1 ^= num_2    num_2 ^= num_1    num_1 ^= num_2    return num_1, num_2'''证明很简单，我们只需要明白异或运算满足下面规律：0^a = a;a^a = 0;a^b^c = a^c^b;'''

又跑题了？

class Solution(object):    def getSum(self, a, b):        """        :type a: int        :type b: int        :rtype: int        """        mask = 0xFFFFFFFF        while b:            a,b=(a^b)&mask,((a&b)<<1)&mask              return a if a <= 0x7FFFFFFF else ~(a ^ mask)

不&mask就会出错

答案

那些用sum的我就不提了

class Solution(object):    def getSum(self, a, b):        """        :type a: int        :type b: int        :rtype: int        """        # 32 bits integer max        MAX = 0x7FFFFFFF        # 32 bits interger min        MIN = 0x80000000        # mask to get last 32 bits        mask = 0xFFFFFFFF        while b != 0:            # ^ get different bits and & gets double 1s, << moves carry            a, b = (a ^ b) & mask, ((a & b) << 1) & mask        # if a is negative, get a's 32 bits complement positive first        # then get 32-bit positive's Python complement negative        return a if a <= MAX else ~(a ^ mask)

Ugly Number

题目

Write a program to check whether a given number is an ugly number.

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 6, 8 are ugly while 14 is not ugly since it includes another prime factor 7.

Note that 1 is typically treated as an ugly number.

思路与解答

class Solution(object):    def isUgly(self, num):        """        :type num: int        :rtype: bool        """        if not num:return False        while num%2==0:            num = num/2        while num%3==0:            num = num/3        while num%5==0:            num = num/5             return True if num ==1 else False

之前居然忽略0了，0不ugly哦

答案

for p in 2, 3, 5:    while num % p == 0 < num:        num /= preturn num == 1

return的很好。循环的也很好，去掉0的判断也很好

class Solution(object):    def isUgly(self, num):        """        :type num: int        :rtype: bool        """        #n = (2**30)*(3**20)*(5**13) = 4570198050078720000000000000L        return False if num < 1 or (4570198050078720000000000000L)%num != 0 else True

巧妙的数学方法
这只是一个数学trick.ex：
如果k = 2 ^ n
则2 ^ 30％k == 0

Two Sum

题目

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:
Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

思路与解答

怎么感觉这题我做过

class Solution(object):    def twoSum(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: List[int]        """        for k,v in enumerate(nums):            if (target-v) in nums:                return [k,nums.index(target-v)]

哦，不能重复使用

class Solution(object):    def twoSum(self, nums, target):        l=[]        for k,v in enumerate(nums):            if (target-v) in l:                return [l.index(target-v),k]            l.append(v)

虽然512ms，40%，但是，不可接受

答案

我看其它答案就是把列表换成了字典，于是我就改了一下

class Solution(object):    def twoSum(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: List[int]        """        l={}        for k,v in enumerate(nums):            if (target-v) in l:                return [l[target-v],k]            l[v] = k

36ms，哇哦噢噢噢噢，爆炸
在字典里查值只需要O(1)

Single Number

题目

Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

思路与解答

想起了异或，写写试试

        return reduce(operator.xor,nums)

so easy!

答案

def singleNumber1(self, nums):    dic = {}    for num in nums:        dic[num] = dic.get(num, 0)+1    for key, val in dic.items():        if val == 1:            return keydef singleNumber2(self, nums):    res = 0    for num in nums:        res ^= num    return resdef singleNumber3(self, nums):    return 2*sum(set(nums))-sum(nums)def singleNumber4(self, nums):    return reduce(lambda x, y: x ^ y, nums)def singleNumber(self, nums):    return reduce(operator.xor, nums)

第3种很有想法
第一种好像可以优化下
抱歉，并不能
不对，如果这样的话

class Solution(object):    def singleNumber(self, nums):        dic = {}        for num in nums:            if num in dic:                del dic[num]            else:                dic[num] = 1        for key,v in dic.items():            return key

确实快了一点