Leetcode 算法題03

575. Distribute Candies

輸入一個含偶數個數字的列表，兩個人均勻地分其中的數字，輸出其中能分到最多不同數字的個數

原本的代碼提交的時候報了超時錯誤，應該是迭代太複雜了

超時的代碼：遍歷次數太多

class Solution(object):
    def distributeCandies(self, candies):
        """
        :type candies: List[int]
        :rtype: int
        """
        anslist = []
        for i in candies:
            if i not in anslist:
                anslist.append(i)
            else:
                pass
        if len(anslist) <= len(candies)//2:
            return len(anslist)
        else:
            return len(candies)//2

思考後的代碼：先將輸入排序，依次遍歷一遍就可以了

class Solution(object):
    def distributeCandies(self, candies):
        """
        :type candies: List[int]
        :rtype: int
        """
        candies.sort()
        anslist = [candies[0]]
        for i in candies:
            if i == anslist[-1]:
                continue
            else:
                anslist.append(i)
        if len(anslist) <= (len(candies)//2):  #這個判斷語句可以用min()來代替
            return len(anslist)
        else:
            return len(candies)//2
        

大神的代碼：理解set集合的概念和運用！！，min函數也要熟練

def distributeCandies(self, candies):
    return min(len(candies) / 2, len(set(candies)))

521. Longest Uncommon Subsequence I

輸入兩個字符串，兩個字符串中存在一個最長的子字符串string，string只存在一個字符串不存在另一個字符串中，求輸出這個子字符串的長度不存在則輸出-1

這道題真的很扯，看懂了就是求較大的字符串長度，兩個字符串一樣時輸出-1，這道題踩的人有1000+

我的代碼：

class Solution(object):
    def findLUSlength(self, a, b):
        """
        :type a: str
        :type b: str
        :rtype: int
        """
        return max(len(a),len(b)) if a != b else -1

717. 1-bit and 2-bit Characters

給定一個最後一位爲0的序列，規定只能由11或10或0組成，求安規定劃分後，最後的0是否爲單獨的

Example 1:

Input: 
bits = [1, 0, 0]
Output: True
Explanation: 
The only way to decode it is two-bit character and one-bit character. So the last character is one-bit character.

Example 2:

Input: 
bits = [1, 1, 1, 0]
Output: False
Explanation: 
The only way to decode it is two-bit character and two-bit character. So the last character is NOT one-bit character.

我的代碼，

class Solution(object):
    def isOneBitCharacter(self, bits):
        """
        :type bits: List[int]
        :rtype: bool
        """
        while len(bits) > 1:
            if bits[0] == 1:
                bits.pop(0)
                bits.pop(0)
            elif bits[0] == 0:
                bits.pop(0)        
        return bits == [0]      #做個copy可能會好一點

大神的代碼：比我的長，但是用索引來找感覺比我的好

    if not bits: return False
    n = len(bits)
    
    index = 0
    while index < n:
        if index == n-1 : return True
        if bits[index] == 1: 
            index += 2              
        else: index += 1
    return False

485. Max Consecutive Ones

輸入一個含0和1的序列，求其中連續最多的1的個數

Example 1:

Input: [1,1,0,1,1,1]
Output: 3
Explanation: The first two digits or the last three digits are consecutive 1s.
    The maximum number of consecutive 1s is 3.

我的代碼：想到啥寫啥，反正都是遍歷一遍，做個表

class Solution(object):
    def findMaxConsecutiveOnes(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        list = []
        count = 0
        for i in nums:
            if i == 1:
                count += 1
            else:
                list.append(count)
                count = 0

	list.append(count)

        return max(list)

大神用了單獨的ans代替list，用max函數判斷是否替換：ans = max(ans, cnt)，其實多想想也能想到

136. Single Number

輸入一個序列，其中每個數字出現2次，除了一個數字，找到這個數

我的代碼：

class Solution(object):
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        nums.sort()
        index = 0
        while index+1 < len(nums):
            if nums[index] == nums[index+1]:
                index += 2
            else:
                return nums[index]
        return nums[index]

大神給出五個版本：

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 key

def singleNumber2(self, nums):
    res = 0
    for num in nums:
        res ^= num
    return res
    
def 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)

693. Binary Number with Alternating Bits

給定正整數，檢查它是否具有交替位：即，如果兩個相鄰位總是具有不同的值。

Example 1:

Input: 5
Output: True
Explanation:
The binary representation of 5 is: 101

Example 2:

Input: 7
Output: False
Explanation:
The binary representation of 7 is: 111.

Example 3:

Input: 11
Output: False
Explanation:
The binary representation of 11 is: 1011.

Example 4:

Input: 10
Output: True
Explanation:
The binary representation of 10 is: 1010.

我的代碼

class Solution(object):
    def hasAlternatingBits(self, n):
        """
        :type n: int
        :rtype: bool
        """
        anslist = str(bin(n)[2:]).split('10')
        for i in range(len(anslist)-1):
            if anslist[i] :
                return False
        return anslist[-1] == '' or anslist[-1] == '1'
        

大神給出了很多思路：給出幾個我看得懂的，代碼沒有python版的

1.取消bit位；2.填充bit位；3.用正則表達式；4.用‘00’‘11’in n

292. Nim Game

兩個人移石子，一次移1-3個，輸入石子個數，求第一個是否能贏

很簡單的問題，想清楚就可以了

class Solution(object):
    def canWinNim(self, n):
        """
        :type n: int
        :rtype: bool
        """
        return n % 4 != 0