从今天开始做Leetcode的算法题，以难度排序，由易到难，简单的时候一天多做一些，一周休息一天，使用python语言，期间看裘宗燕的《数据结构与算法 Python语言描述》，主要目的是能熟练使用python并熟悉基本的数据结构，为面试做准备。
461. Hamming Distance
两个整数转化为二进制后，按位不同的个数，输入0<x,y<2^31
Example:
Input: x = 1, y = 4 Output: 2 Explanation: 1 (0 0 0 1) 4 (0 1 0 0) ↑ ↑ The above arrows point to positions where the corresponding bits are different.
我的代码：
class Solution(object):
def hammingDistance(self, x, y):
"""
:type x: int
:type y: int
:rtype: int
"""
d = '00000000000000000000000000000000'
a=bin(x)[2:]
b=bin(y)[2:]
xbin = d[:-len(a)]+a
ybin = d[:-len(b)]+b
distance = 0
for i in range(32):
if xbin[i] != ybin[i]:
distance += 1
return distanceclass Solution(object):
def hammingDistance(self, x, y):
"""
:type x: int
:type y: int
:rtype: int
"""
return bin(x^y).count('1')初始有一个机器人在座标（0，0）的位置，给出一个移动顺序的序列，如果机器人回到原点则输出True，移动指令分别为：'R','L','U','D'
Example 1:
Input: "UD" Output: true
Example 2:
Input: "LL" Output: false我的代码：
本来想初始化一个座标并进行移动计算，后来想到上面第一题大神的回答，稍微想了一下。
class Solution(object):
def judgeCircle(self, moves):
"""
:type moves: str
:rtype: bool
"""
return moves.count('R') == moves.count('L') and moves.count('U') == moves.count('D')617. Merge Two Binary Trees
树的合并
Example 1:
Input: Tree 1 Tree 2 1 2 / \ / \ 3 2 1 3 / \ \ 5 4 7 Output: Merged tree: 3 / \ 4 5 / \ \ 5 4 7这道题因为完全没接触过树的知识，想了很久也没做出来，基础实在太差了，就看了一下讨论，做了出来，要理解递归与self的意义
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def mergeTrees(self, t1, t2):
"""
:type t1: TreeNode
:type t2: TreeNode
:rtype: TreeNode
"""
if t1 == None and t2 == None:
return None
elif t1 == None:
return t2
elif t2 == None:
return t1
else:
root = TreeNode(t1.val+t2.val)
root.left = self.mergeTrees(t1.left,t2.left)
root.right = self.mergeTrees(t1.right,t2.right)
return root561. Array Partition I
给出一个列表，两两分组为(a1,b1),(a2,b2)...使min(ai,bi)的和尽可能大
Example 1:
Input: [1,4,3,2] Output: 4 Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).我的代码：总之就是将列表排序后隔一个取一个数累加
class Solution:
def arrayPairSum(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
nums.sort()
ans = 0
for i in range(len(nums)//2):
ans += nums[i*2]
return ansclass Solution(object):
def arrayPairSum(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
return sum(sorted(nums)[::2])476. Number Complement
给一个数，转为二进制后按位取反，输出结果的十进制
我的代码：replace那里忘记赋值了，以为和sort()函数一样用了就自动赋值了
class Solution(object):
def findComplement(self, num):
"""
:type num: int
:rtype: int
"""
string = str(bin(num))[2:]
string = string.replace('0','a')
string = string.replace('1','0')
string = string.replace('a','1')
return int(string,2)class Solution(object):
def findComplement(self, num):
i = 1
while i <= num:
i = i << 1
return (i - 1) ^ num500. Keyboard Row
给含多个单词的列表，排除其中所有字母不在键盘一行内的单词
Example 1:
Input: ["Hello", "Alaska", "Dad", "Peace"] Output: ["Alaska", "Dad"]我的代码：
class Solution(object):
def findWords(self, words):
"""
:type words: List[str]
:rtype: List[str]
"""
line=['QWERTYUIOP','ASDFGHJKL','ZXCVBNM']
ans = list(words)
for word in words:
flag = (word[0].upper() in line[0])+2*(word[0].upper() in line[1])
+3*(word[0].upper() in line[2])
for i in word.upper():
if i not in line[flag-1]:
ans.pop(ans.index(word))
break
return ans
def findWords(self, words):
return filter(re.compile('(?i)([qwertyuiop]*|[asdfghjkl]*|[zxcvbnm]*)$').match, words) #(?i) 表示所在位置右侧的表达式开启忽略大小写模式557. Reverse Words in a String III
输入一个句子，将句子中每个单词的字幕顺序置反并输出
Example 1:
Input: "Let's take LeetCode contest" Output: "s'teL ekat edoCteeL tsetnoc"我的代码：
class Solution(object):
def reverseWords(self, s):
"""
:type s: str
:rtype: str
"""
wordlist=s.split()
anslist=[]
ans=""
for word in wordlist:
anslist.append(word[::-1])
for i in anslist:
ans += ' '
ans += i
return ans[1:]
def reverseWords(self, s):
return ' '.join(s.split()[::-1])[::-1]