比较字符串（包含以及变位词）

Two Strings Are Anagrams

Description

Write a method anagram(s,t) to decide if two strings are anagrams or not.

What is Anagram?

• Two strings are anagram if they can be the same after change the order of characters.

Example

Given s = "abcd", t = "dcab", return true.
Given s = "ab", t = "ab", return true.
Given s = "ab", t = "ac", return false.

solution：

class Solution:
"""
@param s: The first string
@param b: The second string
@return true or false
"""
def countchar(self,ch):
result = {}
for item in ch:
ikey = ord(item)
if result.has_key(ikey):
result[ikey] +=1
else:
result.setdefault(ikey,0)
return result

def anagram(self, s, t):
# write your code here
dicS = self.countchar(s)
dicT = self.countchar(t)
if not cmp(dicS,dicT):
return True
else:
return False

Compare Strings

Description

Compare two strings A and B, determine whether A contains all of the characters in B.

The characters in string A and B are all Upper Case letters.

Notice

The characters of B in A are not necessary continuous or ordered.

Example

For A = "ABCD", B = "ACD", return true.

For A = "ABCD", B = "AABC", return false.

solution

class Solution:
"""
@param A : A string includes Upper Case letters
@param B : A string includes Upper Case letters
@return : if string A contains all of the characters in B return True else return False
"""
# used a list with 26 iterm to save the number of each alphabet

def compareStrings(self, A, B):
# write your code here
if B is None:
return True
elif A is None:
return False

CntA = self.countString(A)
CntB = self.countString(B)

for order in range(ord("Z")-ord("A") + 1):
if CntA[order] < CntB[order]:
return False

return True

def countString(self,string):
count = []
for item in range(ord("Z")-ord("A") + 1):
count.append(0)
for item in string:
count[ord(item) - ord("A")] +=1
return count