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‎leetcode/009.回文数.py

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class Solution(object):
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def isPalindrome(self, x):
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return str(x) == str(x)[::-1]

‎leetcode/069.x 的平方根.py

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# 方法一：数学函数
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class Solution(object):
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def mySqrt(self, x):
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return int(math.sqrt(x))
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# 方法二：1/2次方取整
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class Solution(object):
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def mySqrt(self, x):
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return int(x**(1/2.0))

‎leetcode/136.只出现一次的数字.py

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class Solution(object):
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def singleNumber(self, nums):
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d = collections.Counter(nums)
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res = [k for k in d if d[k]==1]
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return res[0]

‎leetcode/155.最小栈.py

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class MinStack(object):
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def __init__(self):
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self.stack = []
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def push(self, x):
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self.stack.append(x)
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def pop(self):
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self.stack.pop()
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def top(self):
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return self.stack[-1]
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def getMin(self):
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return min(self.stack)

‎leetcode/171.Excel表列序号.py

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# 26进制转10进制。正序
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class Solution(object):
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def titleToNumber(self, s):
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res = 0
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for i in range(len(s)):
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res += (26**(len(s)-i-1)) * (ord(s[i])-ord('A')+1)
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return int(res)
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# 逆序
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class Solution(object):
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def titleToNumber(self, s):
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S = list(s)
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S.reverse()
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res = 0
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for i, v in enumerate(S):
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res += (26**i) * (ord(v)-ord('A')+1)
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return res

‎leetcode/172.阶乘后的零.py

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# 找规律，将阶乘中的数分解成因子，有1个5则可以和前面分解出来的2搭配 2*5=10，则末尾就会有1个0
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class Solution(object):
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def trailingZeroes(self, n):
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res = 0
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while n > 0:
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n /= 5
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res += n
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return res

‎leetcode/190.颠倒二进制位.py

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# 方法一：将无符号整数转为二进制，用0补满32位，反转后将二进制格式转为十进制
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class Solution:
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def reverseBits(self, n):
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return int(bin(n)[2:].zfill(32)[::-1], 2)
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# 方法二：位运算，res每次左移加上n的最后一位，然后n右移去掉最后一位
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class Solution:
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def reverseBits(self, n):
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res = 0
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for _ in range(32):
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res = (res<<1) + n%2
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n >>= 1
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return res

‎leetcode/191.位1的个数.py

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# 输入是一个无符号整数，要先转化为二进制形式，再转为字符串形式统计1的个数
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class Solution(object):
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def hammingWeight(self, n):
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return bin(n).count('1')

‎leetcode/202.快乐数.py

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# 所有不快乐数的数位平方和计算，最后都会进入 4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4 的循环中
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class Solution(object):
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def isHappy(self, n):
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while n != 1:
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n = sum(map(lambda x: x**2, map(int, str(n))))
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if n == 4:
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return False
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return True

‎leetcode/204.计数质数.py

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# 厄拉多塞筛法：先将 2~n 的各个数放入表中，然后在2的上面画一个圆圈，然后划去2的其他倍数；
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# 第一个既未画圈又没有被划去的数是3，将它画圈，再划去3的其他倍数；
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# 现在既未画圈又没有被划去的第一个数 是5，将它画圈，并划去5的其他倍数……
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# 依次类推，一直到所有小于或等于 n 的各数都画了圈或划去为止。
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# 这时，表中画了圈的以及未划去的那些数正好就是小于 n 的素数。
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class Solution(object):
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def countPrimes(self, n):
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if n < 3:
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return 0
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primes = [1]*n
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primes[0] = primes[1] = 0
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for i in range(2, int(n**0.5)+1):
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if primes[i]:
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primes[i*i:n:i] = [0]*(len(primes[i*i:n:i]))
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return sum(primes)

‎leetcode/205.同构字符串.py

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# 方法一：若同构则查找字符的索引相同
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class Solution(object):
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def isIsomorphic(self, s, t):
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for i in range(len(s)):
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if s.find(s[i]) != t.find(t[i]):
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return False
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return True
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# 方法二：判断两字符串去重 与 压缩再去重的长度是否相同
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class Solution(object):
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def isIsomorphic(self, s, t):
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if len(s) != len(t):
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return False
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return len(set(zip(s, t))) == len(set(s)) == len(set(t))

‎leetcode/225.用队列实现栈.py

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class MyStack(object):
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def __init__(self):
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self.stack = []
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def push(self, x):
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self.stack.append(x)
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def pop(self):
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return self.stack.pop()
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def top(self):
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return self.stack[-1]
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def empty(self):
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return self.stack == []

‎leetcode/231.2的幂.py

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# 因子可全由2组成，除尽2判断最终结果是否等于1
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class Solution(object):
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def isPowerOfTwo(self, n):
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if n==0:
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return False
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while n%2 == 0:
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n /= 2
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return n == 1

‎leetcode/232.用栈实现队列.py

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class MyQueue(object):
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def __init__(self):
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self.queue = []
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def push(self, x):
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self.queue.append(x)
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def pop(self):
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return self.queue.pop(0)
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def peek(self):
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return self.queue[0]
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def empty(self):
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return self.queue == []

‎leetcode/258.各位相加.py

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# 将字符串拆成当个数字相加，判断结果长度是否大于1，循环此过程
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class Solution(object):
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def addDigits(self, num):
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while len(str(num)) > 1:
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num = sum(map(int, [i for i in str(num)]))
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return num

‎leetcode/263.丑数.py

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# 丑数只包含质因数 2, 3, 5，将质因数除尽看结果是否为1
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class Solution(object):
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def isUgly(self, num):
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if num <= 0:
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return False
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while num%2 == 0:
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num /= 2
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while num%3 == 0:
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num /= 3
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while num%5 == 0:
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num /= 5
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return num == 1

‎leetcode/278.第一个错误的版本.py

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# 二分法，取中间版本进行判断，若是好版本则将左起始点设为中间版本之后，若是坏版本则将右起始点设为为中间版本
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class Solution(object):
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def firstBadVersion(self, n):
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l, r = 1, n
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while l < r:
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mid = (l+r)/2
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if isBadVersion(mid):
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r = mid
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else:
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l = mid + 1
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return l

‎leetcode/290.单词模式.py

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# 判断字符串、列表去重 与 压缩再去重的长度是否相同
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class Solution(object):
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def wordPattern(self, pattern, str):
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s = str.split()
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if len(pattern) != len(s):
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return False
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return len(set(zip(pattern, s))) == len(set(pattern)) == len(set(s))

‎leetcode/292.Nim游戏.py

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# 只要给对手留下4的整数倍个石子就能赢，否则就输
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class Solution(object):
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def canWinNim(self, n):
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return n%4 != 0

‎leetcode/326.3的幂.py

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# 因子可全由3组成，除尽3判断最终结果是否等于1
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class Solution(object):
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def isPowerOfThree(self, n):
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if n==0:
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return False
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while n%3 == 0:
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n /= 3
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return n == 1

‎leetcode/342.4的幂.py

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# 因子可全由4组成，除尽4判断最终结果是否等于1
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class Solution(object):
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def isPowerOfFour(self, num):
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if num==0:
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return False
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while num%4 == 0:
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num /= 4
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return num == 1

‎leetcode/367.有效的完全平方数.py

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# 方法一：直接平方根
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class Solution(object):
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def isPerfectSquare(self, num):
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return (num**0.5) == int(num**0.5)
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# 方法二：二分法，查找是否存在一个数的平方为给定的正整数
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class Solution(object):
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def isPerfectSquare(self, num):
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l, r = 0, num
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while l <= r:
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mid = (l+r)/2
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square = mid**2
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if square > num:
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r = mid - 1
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elif square < num:
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l = mid + 1
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else:
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return True
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return False

‎leetcode/371.两整数之和.py

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class Solution(object):
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def getSum(self, a, b):
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return sum([a, b])

‎leetcode/374.猜数字大小.py

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# 根据返回结果重新定位左右区间
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class Solution(object):
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def guessNumber(self, n):
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l, r = 1, n
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while l <= r:
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mid = (l+r)/2
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if guess(mid) == -1:
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r = mid - 1
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elif guess(mid) == 1:
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l = mid + 1
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else:
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return mid

‎leetcode/389.找不同.py

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# 方法一：遍历s，在t中移除遍历到的元素，最后剩下一个就是添加的字母
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class Solution(object):
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def findTheDifference(self, s, t):
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t = list(t)
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for i in s:
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t.remove(i)
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return t[0]
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# 方法二：遍历t，统计t与s中该字符的个数，若不相同则为添加的字母
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class Solution(object):
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def findTheDifference(self, s, t):
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for i in t:
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if t.count(i) != s.count(i):
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return i

‎leetcode/409.最长回文串.py

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# 统计字符个数存放在字典中，遍历字典。字符个数为偶数则直接累加。
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# 个数为奇数则减1后累加，由于回文串只能有一个字符是奇次数，故若有字符个数是奇数最后要再加1
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class Solution(object):
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def longestPalindrome(self, s):
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d = collections.Counter(s)
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length = k = 0
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for key in d:
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if d[key]%2 == 0:
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length += d[key]
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else:
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k = 1
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length += d[key]-1
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return length + k

‎leetcode/412.Fizz Buzz.py

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class Solution(object):
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def fizzBuzz(self, n):
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res = []
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for i in range(1, n+1):
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if i%3==0 and i%5==0:
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res.append('FizzBuzz')
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elif i%3==0:
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res.append('Fizz')
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elif i%5==0:
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res.append('Buzz')
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else:
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res.append(str(i))
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return res

‎leetcode/415.字符串相加.py

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# eval()执行字符串表达式
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class Solution(object):
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def addStrings(self, num1, num2):
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return str(eval(num1) + eval(num2))

‎leetcode/441.排列硬币.py

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# 每增加一行，当前总硬币数增加该行的行号，当前总硬币数小于等于n时循环，大于时则上一行的行号即为所求总行数
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class Solution(object):
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def arrangeCoins(self, n):
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i = nums = 0
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while nums <= n:
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i += 1
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nums += i
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return i-1

‎leetcode/453.最小移动次数使数组元素相等.py

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# 方法一：每次移动可以使n-1个元素增加1，即使最大元素减1，循环达到所有元素值都为最小元素值
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class Solution(object):
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def minMoves(self, nums):
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min_v = min(nums)
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count = 0
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for num in nums:
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count += num-min_v
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return count
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# 方法二：即方法一同样思路另一种写法
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class Solution(object):
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def minMoves(self, nums):
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return sum(nums) - min(nums) * len(nums)

‎leetcode/455.分发饼干.py

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# 用两个指针指向两个排序后数组的首部，比较两个指针指向的元素。若饼干尺寸满足胃口值则两指针各前进一步，否则饼干指针前进一步
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class Solution(object):
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def findContentChildren(self, g, s):
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g.sort()
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s.sort()
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len_g, len_s = len(g), len(s)
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i = j = 0
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while i<len_g and j<len_s:
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if g[i] <= s[j]:
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i += 1
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j += 1
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else:
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j += 1
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return i

‎leetcode/461.汉明距离.py

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# 异或，其二进制同0异1，再将十进制的异或结果转为二进制，求1的个数
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class Solution(object):
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def hammingDistance(self, x, y):
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return bin(x^y).count('1')

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