标签：string 测试 别人 ide rect pre false detail character
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
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?
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
1 ds,dt = {},{} 2 for w in s: 3 ds[w] = ds.get(w,0)+1 4 for w in t: 5 dt[w] = dt.get(w,0)+1 6 return ds == dt
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 d
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.
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]
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
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.
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
emmmm为啥我要去用指针呢？
1 rev = s[::-1] 2 if s == rev: return True 3 l = len(s) 4 for i in xrange(l): 5 if s[i] != rev[i]: 6 return s[i:l-i-1] == rev[i+1:l-i] or rev[i:l-i-1] == s[i+1:l-i] 7 return False
差不多的方案
1 def validPalindrome(self, s): 2 i = 0 3 while i < len(s) / 2 and s[i] == s[-(i + 1)]: i += 1 4 s = s[i:len(s) - i] 5 return s[1:] == s[1:][::-1] or s[:-1] == s[:-1][::-1]
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.
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
1 stack=[] 2 d={")":"(","}":"{","]":"["} 3 for n in s: 4 if n in d.values(): 5 stack.append(n) 6 elif n in d.keys(): 7 if not stack:return False 8 x = stack.pop() 9 if x != d[n]: 10 return False 11 else: 12 return False 13 return stack == []
速度还行
差不多嘛（但是比我的短）
1 stack = [] 2 pairs = {‘(‘: ‘)‘, ‘{‘: ‘}‘, ‘[‘: ‘]‘} 3 for char in s: 4 if char in pairs: 5 stack.append(pairs[char]) 6 else: 7 if len(stack) == 0 or stack.pop() != char: 8 return False 9 return not stack
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
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
1 n=1 2 while num > 0: 3 num -= n+n-1 4 n += 1 5 return num == 0
别人有更快的，估计是方法不一样
emmm就是之前的某个公式，居然比我的快。
1 def isPerfectSquare(self, num): 2 x = num 3 r = x 4 while r*r > x: 5 r = (r + x/r) / 2 6 return r*r == x
标签：string 测试 别人 ide rect pre false detail character
原文地址：https://www.cnblogs.com/huohuohuo1/p/9286375.html
