0610

Author: princewen

easy/二进制/371_Sum of Two Integers.py

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+"""
+
+Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.
+
+Example:
+Given a = 1 and b = 2, return 3.
+
+Credits:
+Special thanks to @fujiaozhu for adding this problem and creating all test cases.
+
+"""
+
+"""
+思路：强大的位运算，不能用+ and -，考虑位运算，按位加
+那么 1+1=10相当于 按位与再左移1位，1+0=1相当于按位异或，所以 按照这个思路进行迭代
+
+"""
+"""但是，这个对python不适用！，当输入负数时，会超时"""
+class Solution(object):
+    def getSum(self, a, b):
+        """
+        :type a: int
+        :type b: int
+        :rtype: int
+        """
+        if a==0:
+            return b
+        if b==0:
+            return a
+        while b:
+            carry = a & b
+            a = a ^ b
+            b = carry << 1
+        return a
+
+
+"""python solution"""
+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
+        """有疑问,python把有符号数当作无符号数处理"""
+        return a if a <= MAX else ~(a ^ mask)

easy/二进制/A summary: how to use bit manipulation to solve problems easily and efficiently | LeetCode Discuss.pdf

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+def getSum( 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)
+
+mask = 0xFFFFFFFF
+print (getSum(-1,-2))
+print (~(4294967293^ mask))

easy/二进制/test.py

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+"""
+
+Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false.
+
+Each letter in the magazine string can only be used once in your ransom note.
+
+Note:
+You may assume that both strings contain only lowercase letters.
+
+canConstruct("a", "b") -> false
+canConstruct("aa", "ab") -> false
+canConstruct("aa", "aab") -> true
+Subscribe to see which companies asked this question.
+
+"""
+"""
+solution:利用counter的计数功能
+"""
+
+import collections
+
+class Solution(object):
+    def canConstruct(self, ransomNote, magazine):
+        """
+        :type ransomNote: str
+        :type magazine: str
+        :rtype: bool
+        """
+        return not collections.Counter(ransomNote) - collections.Counter(magazine)

easy/字符串/383_Ransom Note.py

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+"""
+
+Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.
+
+Examples:
+
+s = "leetcode"
+return 0.
+
+s = "loveleetcode",
+return 2.
+Note: You may assume the string contain only lowercase letters.
+
+"""
+import string
+
+class Solution(object):
+    def firstUniqChar(self, s):
+        """
+        :type s: str
+        :rtype: int
+        """
+        return min([s.find(c) for c in string.ascii_lowercase if s.count(c)==1] or [-1])

easy/字符串/387_First Unique Character in a String.py

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+"""
+
+Given two strings s and t which consist of only lowercase letters.
+
+String t is generated by random shuffling string s and then add one more letter at a random position.
+
+Find the letter that was added in t.
+
+Example:
+
+Input:
+s = "abcd"
+t = "abcde"
+
+Output:
+e
+
+Explanation:
+'e' is the letter that was added.
+
+"""
+
+"""my solution：两个字符串只有一个值不同，类似于之前的一道题，一个数组中所有数字都出现了两次，只有一个数出现了一次，对数组中所有的数进行一次异或操作即可"""
+
+class Solution(object):
+    def findTheDifference(self, s, t):
+        """
+        :type s: str
+        :type t: str
+        :rtype: str
+        """
+        c = s+t
+        res = 0
+        for single_char in c:
+            res = res ^ ord(single_char)
+        return chr(res)
+
+"""可以写的更精炼"""
+
+
+class Solution(object):
+    def findTheDifference(self, s, t):
+        """
+        :type s: str
+        :type t: str
+        :rtype: str
+        """
+        return chr(reduce(operator.xor, map(ord, (s + t))))

easy/字符串/389_Find the Difference.py

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+"""
+
+Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.
+
+This is case sensitive, for example "Aa" is not considered a palindrome here.
+
+Note:
+Assume the length of given string will not exceed 1,010.
+
+Example:
+
+Input:
+"abccccdd"
+
+Output:
+7
+
+Explanation:
+One longest palindrome that can be built is "dccaccd", whose length is 7.
+
+"""
+
+"""
+my solution：结果分为三部分，计数数量为正数的，随后计数为单数的-1，最后，如果有单数元素存在，长度肯定可以加1
+这里注意一个testcase "ccc"，应该返回3，瑞后的max要注意，只要有计数为单数的字符，都要加上1，而不是只有计数为1的才行
+
+"""
+
+import collections
+class Solution(object):
+    def longestPalindrome(self, s):
+        """
+        :type s: str
+        :rtype: int
+        """
+        t = collections.Counter(s)
+        return sum([t[x] for x in t if t[x] % 2 == 0]) + sum([t[x] - 1 for x in t if  t[x] % 2 == 1]) + max(
+            [1 for x in t if t[x] % 2 == 1] or [0])

easy/字符串/409_Longest Palindrome.py

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+"""
+
+Given two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2.
+
+Note:
+
+The length of both num1 and num2 is < 5100.
+Both num1 and num2 contains only digits 0-9.
+Both num1 and num2 does not contain any leading zero.
+You must not use any built-in BigInteger library or convert the inputs to integer directly.
+
+"""
+
+"""my solution
+从最后一位往前加，其实比较繁琐
+"""
+
+
+class Solution(object):
+    def addStrings(self, num1, num2):
+        """
+        :type num1: str
+        :type num2: str
+        :rtype: str
+        """
+        extra = 0
+        sum = ''
+        for i in range(1,min(len(num1),len(num2))+1):
+            sum = str((ord(num1[-i])+ord(num2[-i])-96+extra)%10)+sum
+            extra = (ord(num1[-i])+ord(num2[-i])+extra-96)/10
+        if len(num1) > i:
+            for j in range(i+1,len(num1)+1):
+                sum = str((ord(num1[-j])-48+extra)%10)+sum
+                extra = (ord(num1[-j])+extra-48)/10
+        if len(num2) > i:
+            for j in range(i+1,len(num2)+1):
+                sum = str((ord(num2[-j])-48+extra)%10)+sum
+                extra = (ord(num2[-j])+extra-48)/10
+        if extra>0:
+            sum = str(extra)+sum
+        return sum

easy/字符串/415_Add Strings.py

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+"""
+
+Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters.
+
+Please note that the string does not contain any non-printable characters.
+
+Example:
+
+Input: "Hello, my name is John"
+Output: 5
+
+
+"""
+
+"""要考虑连续空格的情形"""
+
+
+class Solution(object):
+    def countSegments(self, s):
+        """
+        :type s: str
+        :rtype: int
+        """
+        if not s:
+            return 0
+        ss = s.split(' ')
+        return len([x for x in ss if x.strip() != ''])
+
+"""其实一行就可以完成！不指定分隔符默认按照空格进行划分，包括连续的空格！"""
+class Solution(object):
+    def countSegments(self, s):
+        """
+        :type s: str
+        :rtype: int
+        """
+        return len(s.split())
+
+

easy/字符串/434_Number of Segments in a String.py

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+"""
+
+Given a positive integer num, write a function which returns True if num is a perfect square else False.
+
+Note: Do not use any built-in library function such as sqrt.
+
+Example 1:
+
+Input: 16
+Returns: True
+Example 2:
+
+Input: 14
+Returns: False
+
+"""
+
+"""my solution
+二分搜索的方法
+"""
+
+class Solution(object):
+    def isPerfectSquare(self, num):
+        """
+        :type num: int
+        :rtype: bool
+        """
+        if num < 0:
+            return False
+        left = 0
+        right = num
+        while left <= right:
+            mid = (right-left)/2+left
+            if mid * mid == num:
+                return True
+            elif mid * mid < num:
+                left = mid + 1
+            else:
+                right = mid - 1
+        return False
+
+"""other solution
+使用牛顿方法解:http://blog.csdn.net/hyc__/article/details/41117009
+"""
+
+class Solution(object):
+    def isPerfectSquare(self, num):
+        """
+        :type num: int
+        :rtype: bool
+        """
+        r = num
+        while r*r > num:
+            r = (r + num/r) / 2
+        return r*r == num
+