0622

Author: princewen

sort_by_leetcode/math/__init__.py

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+"""
+
+Given a positive integer, return its corresponding column title as appear in an Excel sheet.
+
+For example:
+
+    1 -> A
+    2 -> B
+    3 -> C
+    ...
+    26 -> Z
+    27 -> AA
+    28 -> AB 
+Credits:
+Special thanks to @ifanchu for adding this problem and creating all test cases.
+
+"""
+
+"""采用除余结合的思想，但是要考虑26%26 为0，所以我们可以先对n-1，再除以26，再加上'A'，"""
+
+
+class Solution(object):
+    def convertToTitle(self, n):
+        """
+        :type n: int
+        :rtype: str
+        """
+        res = ''
+        while n:
+            res = chr((n - 1) % 26 + ord('A')) + res
+            n = (n - 1) / 26
+        return res

sort_by_leetcode/math/easy/168. Excel Sheet Column Title.py

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+"""
+
+Related to question Excel Sheet Column Title
+
+Given a column title as appear in an Excel sheet, return its corresponding column number.
+
+For example:
+
+    A -> 1
+    B -> 2
+    C -> 3
+    ...
+    Z -> 26
+    AA -> 27
+    AB -> 28 
+
+"""
+"""168题逆向思维"""
+class Solution(object):
+    def titleToNumber(self, s):
+        """
+        :type s: str
+        :rtype: int
+        """
+        res = 0
+        for num in s:
+            res = res * 26 + (ord(num)-ord('A')+1)
+        return res

sort_by_leetcode/math/easy/171. Excel Sheet Column Number.py

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+"""
+
+Given an integer n, return the number of trailing zeroes in n!.
+
+Note: Your solution should be in logarithmic time complexity.
+
+"""
+"""直接使用数学公式即可:0的个数= n/5 + n/(5^2)+.......
+"""
+class Solution(object):
+    def trailingZeroes(self, n):
+        zeroCnt = 0
+        while n > 0:
+            n = n/ 5
+            zeroCnt += n
+
+        return zeroCnt

sort_by_leetcode/math/easy/172. Factorial Trailing Zeroes.py

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+"""
+
+Write an algorithm to determine if a number is "happy".
+
+A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.
+
+Example: 19 is a happy number
+
+12 + 92 = 82
+82 + 22 = 68
+62 + 82 = 100
+12 + 02 + 02 = 1
+Credits:
+Special thanks to @mithmatt and @ts for adding this problem and creating all test cases.
+
+"""
+
+"""解法1，用列表保存出现过的数，当出现重复的数字的时候，说明出现了循环，循环有两种情况，一种不是hayyp数，循环的数必不是1，如果是happy数，那么会出现1的不断循环"""
+class Solution(object):
+    def isHappy(self, n):
+        """
+        :type n: int
+        :rtype: bool
+        """
+        c = set()
+        while not n in c:
+            c.add(n)
+            n = sum([int(x) ** 2 for x in str(n)])
+        return n==1
+
+"""解法2，使用追赶法，快的指针每次前进两步，慢的指针每次只前进一步，当快的指针追上慢的指针即二者数字相同时，同样说明出现了
+循环，情况同上"""
+class Solution(object):
+    def isHappy(self, n):
+        """
+        :type n: int
+        :rtype: bool
+        """
+        slow = n
+        quick = sum([int(x) ** 2 for x in str(n)])
+        while quick != slow:
+            quick = sum([int(x) ** 2 for x in str(quick)])
+            quick = sum([int(x) ** 2 for x in str(quick)])
+            slow = sum([int(x) ** 2 for x in str(slow)])
+        return slow == 1

sort_by_leetcode/math/easy/202. Happy Number.py

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+"""
+
+Description:
+
+Count the number of prime numbers less than a non-negative number, n.
+
+"""
+
+"""统计比n小的质数有多少，首先设置一个数组，全部质为true，然后遍历2-sqrt（n）的数，把他们的倍数所在的数组位置
+置为True这里为了避免重复，从i*i的位置开始，而不是从i*2的位置开始，因为i*2，i*3，i*n-1其实已经置为false了
+"""
+class Solution(object):
+    def countPrimes(self, n):
+        """
+        :type n: int
+        :rtype: int
+        """
+        if n < 3:
+            return 0
+        res = [True] * n
+        res[0] = res[1] = False
+        for i in range(2,int(math.sqrt(n)) + 1):
+            res[i*i:n:i] = [False] * len(res[i*i:n:i])
+        return sum(res)

sort_by_leetcode/math/easy/204. Count Primes.py

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+"""
+
+Given a non-negative integer represented as a non-empty array of digits, plus one to the integer.
+
+You may assume the integer do not contain any leading zero, except the number 0 itself.
+
+The digits are stored such that the most significant digit is at the head of the list.
+
+"""
+"""利用了str和int的互相转换"""
+class Solution(object):
+    def plusOne(self, digits):
+        """
+        :type digits: List[int]
+        :rtype: List[int]
+        """
+        sum = 0
+        for i in digits:
+            sum = sum * 10 + i
+        return [int(x) for x in str(sum+1)]

sort_by_leetcode/math/easy/66. Plus One.py

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+"""
+
+Given two binary strings, return their sum (also a binary string).
+
+For example,
+a = "11"
+b = "1"
+Return "100".
+
+"""
+
+"""
+采用递归的思路，从末位开始相加，要分五种情况进行讨论
+1、a为空，返回b
+2、b为空，返回a
+3、a和b的末位都是1，递归的同时，末位添0，同时前面要再和'1'想加
+4、a和b的末位都是0，递归的同时，末位添0
+5、a和b的末位有一个为1，递归的同时，末位添1
+"""
+
+
+class Solution(object):
+    def addBinary(self, a, b):
+        """
+        :type a: str
+        :type b: str
+        :rtype: str
+        """
+        if a == '':
+            return b
+        if b == '':
+            return a
+        if a[-1] == '1' and b[-1] == '1':
+            return self.addBinary(self.addBinary(a[:-1], b[:-1]), '1') + '0'
+        elif a[-1] == '0' and b[-1] == '0':
+            return self.addBinary(a[:-1], b[:-1]) + '0'
+        else:
+            return self.addBinary(a[:-1], b[:-1]) + '1'

sort_by_leetcode/math/easy/67. Add Binary.py

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+"""
+
+Implement int sqrt(int x).
+
+Compute and return the square root of x.
+
+"""
+"""采用二分搜索的思路"""
+
+class Solution(object):
+    def mySqrt(self, x):
+        """
+        :type x: int
+        :rtype: int
+        """
+        if x<=0:
+            return 0
+        else:
+            left = 1
+            right = x
+            while left <= right:
+                mid = (left+right) / 2
+                a1 = mid*mid - x
+                a2 = (mid-1)*(mid-1) -x
+                if a1*a2 <= 0:
+                    break
+                elif a1 * a2 >0 and a1<0:
+                    left = mid+1
+                else:
+                    right = mid-1
+            if a1 == 0:
+                return mid
+            else:
+                return mid-1

sort_by_leetcode/math/easy/69. Sqrt(x).py

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+"""
+
+Reverse digits of an integer.
+
+Example1: x = 123, return 321
+Example2: x = -123, return -321
+
+click to show spoilers.
+
+Note:
+The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.
+
+"""
+
+
+"""
+这里用到的方法是经典的除余相结合，得到一个数的最后一位简单的方法就是余10，得到这个数的除最后一位之外的n-1位，简单的方法就是除10
+这里要注意的地方就是负数首先要先得到它的相反数在进行操作，最后再变为负数
+这个题还有一个注意的地方就是别的语言如c++，java中整数有溢出，而python的整数是没有溢出的，所以我们最好手动判断是否溢出
+"""
+
+class Solution(object):
+    def reverse(self, x):
+        """
+        :type x: int
+        :rtype: int
+        """
+        n = x if x > 0 else -x
+        res = 0
+        while n:
+            res = res * 10 + n % 10
+            n = n / 10
+        if res > 0x7fffffff:
+            return 0
+        return res if x > 0 else -res