LintCode_chapter1_section4_anagrams

coding = utf-8

'''
Created on 2015年11月5日

@author: SphinxW
'''
乱序字符串

给出一个字符串数组S，找到其中所有的乱序字符串(Anagram)。如果一个字符串是乱序字符串，那么他存在一个字母集合相同，但顺序不同的字符串也在S中。
您在真实的面试中是否遇到过这个题？
样例

对于字符串数组 ["lint","intl","inlt","code"]

返回 ["lint","inlt","intl"]
注意

所有的字符串都只包含小写字母

class Solution:
    # @param strs: A list of strings
    # @return: A list of strings

    def anagrams(self, strs):
        # write your code here
        countStrs = {}
        result = []
        for index, item in enumerate(strs):
            countItem = self.countChars(item)
            try:
                if index not in countStrs[countItem]:
                    countStrs[countItem].append(index)
            except KeyError:
                countStrs[countItem] = [index, ]
        for key, item in countStrs.items():
            if len(item) > 1:
                for s in item:
                    result.append(strs[s])
        return result

    def anagram(self, s, t):
        countsS = self.countChars(s)
        countsT = self.countChars(t)
        return countsS == countsT

    def countChars(self, stringToCount):
        result = []
        for item in range(ord("z") - ord("A")+1):
            result.append(0)
        for item in stringToCount:
            result[ord(item) - ord("A")] += 1
        return str(result)