從今天開始做Leetcode的算法題，以難度排序，由易到難，簡單的時候一天多做一些，一週休息一天，使用python語言，期間看裘宗燕的《數據結構與算法 Python語言描述》，主要目的是能熟練使用python並熟悉基本的數據結構，爲面試做準備。
461. Hamming Distance
兩個整數轉化爲二進制後，按位不同的個數，輸入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.
我的代碼：
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')初始有一個機器人在座標（0，0）的位置，給出一個移動順序的序列，如果機器人回到原點則輸出True，移動指令分別爲：'R','L','U','D'
Example 1:
Input: "UD" Output: true
Example 2:
Input: "LL" Output: false我的代碼：
本來想初始化一個座標並進行移動計算，後來想到上面第一題大神的回答，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
樹的合併
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這道題因爲完全沒接觸過樹的知識，想了很久也沒做出來，基礎實在太差了，就看了一下討論，做了出來，要理解遞歸與self的意義
# 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
給出一個列表，兩兩分組爲(a1,b1),(a2,b2)...使min(ai,bi)的和儘可能大
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).我的代碼：總之就是將列表排序後隔一個取一個數累加
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
給一個數，轉爲二進制後按位取反，輸出結果的十進制
我的代碼：replace那裏忘記賦值了，以爲和sort()函數一樣用了就自動賦值了
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
給含多個單詞的列表，排除其中所有字母不在鍵盤一行內的單詞
Example 1:
Input: ["Hello", "Alaska", "Dad", "Peace"] Output: ["Alaska", "Dad"]我的代碼：
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) 表示所在位置右側的表達式開啓忽略大小寫模式557. Reverse Words in a String III
輸入一個句子，將句子中每個單詞的字幕順序置反並輸出
Example 1:
Input: "Let's take LeetCode contest" Output: "s'teL ekat edoCteeL tsetnoc"我的代碼：
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]