LeetCode周赛150——1160. Find Words That Can Be Formed by Characters

灰太狼 2023-08-17 15:14 10阅读 0赞

1.4 2800.3800

本来不是很有信心，哦不，是做好了陪跑的准备。因为最后10分了还有错，觉得没戏了，结果哈哈哈，天道好轮回，苍天绕过谁，哦不，死气白咧，必有回响。废话不多说，开始分析下这个仅存的硕果。感觉思路是没毛病的，所以是基本python知识点不熟练导致的语法问题。还是要多读python的代码尤其是大段的，一定会有帮助。老铁，撸起袖子加油干

### 1160. Find Words That Can Be Formed by Characters ###

[My Submissions][] [Back to Contest][]

*  User Accepted:2730
 *  User Tried:2866
 *  Total Accepted:2776
 *  Total Submissions:3902
 *  Difficulty:Easy

You are given an array of strings `words` and a string `chars`.

A string is *good* if it can be formed by characters from `chars` (each character can only be used once).

Return the sum of lengths of all good strings in `words`.

Example 1:

Input: words = ["cat","bt","hat","tree"], chars = "atach"
    Output: 6
    Explanation: 
    The strings that can be formed are "cat" and "hat" so the answer is 3 + 3 = 6.

Example 2:

Input: words = ["hello","world","leetcode"], chars = "welldonehoneyr"
    Output: 10
    Explanation: 
    The strings that can be formed are "hello" and "world" so the answer is 5 + 5 = 10.

Note:

1.  `1 <= words.length <= 1000`
2.  `1 <= words[i].length, chars.length <= 100`
3.  All strings contain lowercase English letters only.

Python version:

BONUS:

1.self代表类的实例，而非类。==只有在类里面会有self出现

同理只有在类的函数里会有self这个多余参数的出现，正常的函数参数声明是不需要self的

2.注意题目给的注释对输入输出变量的描述。The type of the variable limits the function use of this

3.辨析五种基本数据类型，7种运算符类型

https://www.cnblogs.com/snaildev/p/7553087.html

https://www.cnblogs.com/snaildev/p/7544558.html

4.base on 3.different types of parentheses means different initialization type

d=\{\}

d=\[\]

d=()

5.List-->intersection union complement交集并集补集运算

[https://www.cnblogs.com/jlf0103/p/8882896.html][https_www.cnblogs.com_jlf0103_p_8882896.html]

class Solution(object):
    
        def countCharacters(self, words, chars):
            """
            :type words: List[str]
            :type chars: str
            :rtype: int
            """
            ans=0
            Dic=self.Comp(chars)
            for tmp in words:
                Dic1=self.Comp(tmp)
                if self.CompDic(Dic,Dic1):
                    ans+=len(tmp)
            return ans
    
        def CompDic(self,dict, dict1):
            keys = list(set(dict).intersection(set(dict1)))
            if len(keys) != len(dict1):
                return False
            for i in keys:
                if dict1[i] > dict[i]:
                    return False
                else:
                    dict1[i] += 1
            return True
    
        def Comp(self,chars):
            """input string return dict"""
            d = {}  # 给出字典
            for x in chars:
                if x in d:
                    d[x] = d[x] + 1
                else:
                    d[x] = 1
            # print(d)
            return d
    
    
    solu=Solution()
    n=solu.countCharacters(words=["cat","bt","hat","tree"],chars="atach")
    print(n)

不自知

转载于:https://www.cnblogs.com/Marigolci/p/11373039.html

[My Submissions]: https://leetcode.com/contest/weekly-contest-150/problems/find-words-that-can-be-formed-by-characters/submissions/
[Back to Contest]: https://leetcode.com/contest/weekly-contest-150/
[https_www.cnblogs.com_jlf0103_p_8882896.html]: https://www.cnblogs.com/jlf0103/p/8882896.html