注意，答案只是代表是他人写的代码，正确，但不一定能通过测试（比如超时），列举出来只是它们拥有着独到之处，虽然大部分确实比我的好
242. Valid Anagram
题目
Given two strings s and t, write a function to determine if t is an anagram of s.
For example,
s = “anagram”, t = “nagaram”, return true.
s = “rat”, t = “car”, return false.
Note:
You may assume the string contains only lowercase alphabets.
Follow up:
What if the inputs contain unicode characters? How would you adapt your solution to such case?
思路与解答
不是很懂啥意思？
是s和t使用了相同的字符？
set喽
好像还要求字符数相等？
dict呗
ds,dt = {},{}
for w in s:
ds[w] = ds.get(w,0)+1
for w in t:
dt[w] = dt.get(w,0)+1
return ds == dt答案
啊，我也想过用sorted做。。。但是一闪而过又忘记了？
return sorted(s) == sorted(t)return all([s.count(c)==t.count(c) for c in string.ascii_lowercase])return collections.Counter(s)==collections.Counter(t)真是各种一行方案啊
看到有人说一个dict就能解决，想了一下是的
#是我写的
d = {}
for w in s:
d[w] = d.get(w,0)+1
for w in t:
d[w] = d.get(w,0)-1
if not d[w]:del d[w]
return not d125. Valid Palindrome
题目
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
For example,
“A man, a plan, a canal: Panama” is a palindrome.
“race a car” is not a palindrome.
Note:
Have you consider that the string might be empty? This is a good question to ask during an interview.
For the purpose of this problem, we define empty string as valid palindrome.
思路与解答
那个例子举得我有些不懂呢？？？
“A man, a plan, a canal: Panama” is a palindrome.
Why？？？
哦，只要求字母对称就可以了啊
判断是不是字母我记得有个函数来着
n=[i.lower() for i in s if i.isalnum()]
return n == n[::-1]答案
指针方案，没有去考虑这么写（因为毕竟麻烦）
def isPalindrome(self, s):
l, r = 0, len(s)-1
while l < r:
while l < r and not s[l].isalnum():
l += 1
while l <r and not s[r].isalnum():
r -= 1
if s[l].lower() != s[r].lower():
return False
l +=1; r -= 1
return True我也看到和我一样的方法了，不过好像被批评了？
680. Valid Palindrome II
题目
Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome.
Example 1:
Input: “aba”
Output: True
Example 2:
Input: “abca”
Output: True
Explanation: You could delete the character ‘c’.
Note:
The string will only contain lowercase characters a-z. The maximum length of the string is 50000.
思路与解答
卧槽，如果每删一个对比一次。。。感觉会超时的吧
不对，先比较回文，出错再考虑方案
这样就要用到之前的指针方案了
在处理第一个错误那里出现了问题，怎么保证你删的那个是对的呢。。。感觉要完全比较下去
def huiwen(n,f):
l,r = 0, len(n)-1
while l < r:
if n[l]!= n[r]:
if f:
return huiwen(n[l+1:r+1],0) or huiwen(n[l:r],0)
else:
return False
l += 1
r -= 1
return True
return huiwen(s,1)因为要套几遍，所以我直接写个函数了
可惜速度不行啊
答案
emmmm为啥我要去用指针呢？
rev = s[::-1]
if s == rev: return True
l = len(s)
for i in xrange(l):
if s[i] != rev[i]:
return s[i:l-i-1] == rev[i+1:l-i] or rev[i:l-i-1] == s[i+1:l-i]
return False差不多的方案
def validPalindrome(self, s):
i = 0
while i < len(s) / 2 and s[i] == s[-(i + 1)]: i += 1
s = s[i:len(s) - i]
return s[1:] == s[1:][::-1] or s[:-1] == s[:-1][::-1]20. Valid Parentheses
题目
Given a string containing just the characters ‘(‘, ‘)’, ‘{‘, ‘}’, ‘[’ and ‘]’, determine if the input string is valid.
The brackets must close in the correct order, “()” and “()[]{}” are all valid but “(]” and “([)]” are not.
思路与解答
堆栈？
如何将字典里的两个括号关联起来？
不能根据values查找key。。。。
d.items()怎么不对？？
好像可以去掉，后面还有判断的
stack=[]
d={")":"(","}":"{","]":"["}
for n in s:
if n in d.values():
stack.append(n)
elif n in d.keys():
if not stack:return False
x = stack.pop()
if x != d[n]:
return False
else:
return False
return stack == []速度还行
答案
差不多嘛（但是比我的短）
stack = []
pairs = {'(': ')', '{': '}', '[': ']'}
for char in s:
if char in pairs:
stack.append(pairs[char])
else:
if len(stack) == 0 or stack.pop() != char:
return False
return not stack367. Valid Perfect Square
题目
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
思路与解答
意思是这个数是不是其它整数的平方？
感觉需要搜一下判断方法
完全平方数等于1+3+5+7+9+….+2n-1
比暴力版快
n=1
while num > 0:
num -= n+n-1
n += 1
return num == 0别人有更快的，估计是方法不一样
答案
emmm就是之前的某个公式，居然比我的快
def isPerfectSquare(self, num):
x = num
r = x
while r*r > x:
r = (r + x/r) / 2
return r*r == x