• Valid Anagram
  • 題目
  • 思路與解答
  • 答案
• Valid Palindrome
  • 題目
  • 思路與解答
  • 答案
• Valid Palindrome II
  • 題目
  • 思路與解答
  • 答案
• Valid Parentheses
  • 題目
  • 思路與解答
  • 答案
• Valid Perfect Square
• 題目
• 思路與解答
• 答案

註意，答案只是代表是他人寫的代碼，正確，但不一定能通過測試（比如超時），列舉出來只是它們擁有著獨到之處，雖然大部分確實比我的好

1. Valid Anagram

題目

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?

思路與解答

不是很懂啥意思？
是s和t使用了相同的字符？
set嘍
好像還要求字符數相等？
dict唄

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

答案

啊，我也想過用sorted做。。。但是一閃而過又忘記了？

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)

真是各種一行方案啊
看到有人說一個dict就能解決，想了一下是的。

        #是我寫的
        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

2. Valid Palindrome

題目

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.

思路與解答

那個例子舉得我有些不懂呢？？？
“A man, a plan, a canal: Panama” is a palindrome.
Why？？？
哦，只要求字母對稱就可以了啊
判斷是不是字母我記得有個函數來著

        n=[i.lower() for i in s if i.isalnum()]
        return n == n[::-1]

答案

指針方案，沒有去考慮這麽寫（因為畢竟麻煩）

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

3. Valid Palindrome II

題目

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.

思路與解答

臥槽，如果每刪一個對比一次。。。感覺會超時的吧
不對，先比較回文，出錯再考慮方案
這樣就要用到之前的指針方案了
在處理第一個錯誤那裏出現了問題，怎麽保證你刪的那個是對的呢。。。感覺要完全比較下去。

        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)

因為要套幾遍，所以我直接寫個函數了
可惜速度不行啊

答案

emmmm為啥我要去用指針呢？

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]

4. Valid Parentheses

題目

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.

思路與解答

堆棧？
如何將字典裏的兩個括號關聯起來？
不能根據values查找key。。。。
d.items()怎麽不對？？
好像可以去掉，後面還有判斷的

 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 == []

速度還行

答案

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

5. Valid Perfect Square

題目

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

思路與解答

意思是這個數是不是其它整數的平方？
感覺需要搜一下判斷方法
完全平方數等於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

別人有更快的，估計是方法不一樣

答案

emmm就是之前的某個公式，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

五個實用的Python案例題，非常有用！