题目

python 代码

排序比较

def isAnagram(s, t):
    # 排序比较
    if len(s) != len(t):
        return False
    sList = list(s)
    sList.sort()
    tList = list(t)
    tList.sort()
    for i in range(len(s)):
        if sList[i] == tList[i]:
            continue
        else:
            return False
    return True

列表化后删除

def isAnagram(s, t):
    #列表化后删除
    if len(s) != len(t):
        return False
    sList = list(s)
    for word in t:
        try:
            sList.remove(word)
        except ValueError as e:
            return False
    return True

利用字典统计

def isAnagram(s, t):
    dict1={}
    dict2={}
    for i in s:
        if i not in dict1:
            dict1[i] = 1
        else:
            dict1[i] += 1
    for i in t:
        if i not in dict2:
            dict2[i]=1
        else:
            dict2[i]+=1
    if len(dict1)!=len(dict2):
        return False
    else:
        for key,value in dict1.items():
            if key not in dict2:
                return False
            else:
                if value == dict2[key]:
                    continue
                else:
                    return False
    return True

标签：dict1,return,dict2,--,len,else,力扣,242,False
来源： https://blog.csdn.net/weixin_48994268/article/details/117414140