171. 乱序字符串

class Solution:
    """
    @param strs: A list of strings
    @return: A list of strings
    """
    ‘‘‘
    大致思路：
    1.给出一个方法，判断两个字符串是否是乱序相等
    2.初始化res，while strs不为空的时候，内层循环，pop值进行判断，符合写入res，返回
    ‘‘‘
    def anagrams(self, strs):
        res = []
        isEq = False
        while strs != []:
            p = strs.pop(0)
            if p in res:
                continue
            for s in strs:
                if self.isEqual(p,s) == True:
                    res.append(s)
                    isEq = True
            if isEq == True:
                res.append(p)
            isEq = False
        return res

            
    def isEqual(self,s1,s2):
        while s1 != ‘‘:
            p = s1[0]
            if p in s2:
                s2 = s2.replace(p,‘‘,1)
            else:
                return False
            s1 = s1.replace(p,‘‘,1)
        return True if s2 == ‘‘ else False

class Solution:
    """
    @param strs: A list of strings
    @return: A list of strings
    """
    ‘‘‘
    大致思路:
    1.首先给出字典,{sortword:[xx,xx]}
    2.循环字典k，v，如果v的长度大于1，则写入res里面
    ‘‘‘
    def anagrams(self, strs):
        dic = {}
        res = []
        for word in strs:
            sortedword =  ‘‘.join(i for i in sorted(word))
            dic[sortedword] = [word] if sortedword  not in dic.keys() else dic[sortedword] + [word]
        
        #对dic操作
        for k,v in dic.items():
            if len(v) > 1:
                res.extend(v)
        return res