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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)

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