843 Guess the Word

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本文介绍了一种在限定次数内找出隐藏单词的算法。通过分析LeetCode平台上一款互动猜词游戏，提出了一种策略，利用guess函数反馈缩小可能的单词范围，最终找到目标单词。文章详细解释了算法思路，并提供了实现代码。

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1 题目

This problem is an interactive problem new to the LeetCode platform.

We are given a word list of unique words, each word is 6 letters long, and one word in this list is chosen as secret.

You may call master.guess(word) to guess a word. The guessed word should have type string and must be from the original list with 6 lowercase letters.

This function returns an integer type, representing the number of exact matches (value and position) of your guess to the secret word. Also, if your guess is not in the given wordlist, it will return -1 instead.

For each test case, you have 10 guesses to guess the word. At the end of any number of calls, if you have made 10 or less calls to master.guess and at least one of these guesses was the secret, you pass the testcase.

Besides the example test case below, there will be 5 additional test cases, each with 100 words in the word list. The letters of each word in those testcases were chosen independently at random from 'a' to 'z', such that every word in the given word lists is unique.

Example 1:
Input: secret = "acckzz", wordlist = ["acckzz","ccbazz","eiowzz","abcczz"]

Explanation:

master.guess("aaaaaa") returns -1, because "aaaaaa" is not in wordlist.
master.guess("acckzz") returns 6, because "acckzz" is secret and has all 6 matches.
master.guess("ccbazz") returns 3, because "ccbazz" has 3 matches.
master.guess("eiowzz") returns 2, because "eiowzz" has 2 matches.
master.guess("abcczz") returns 4, because "abcczz" has 4 matches.

We made 5 calls to master.guess and one of them was the secret, so we pass the test case.

2 尝试解

2.1 分析

给定一组长度均为6的字符串words，其中一个是谜底。提供函数guess(string&A)，返回A与谜底在相同位置上相同字符的个数，其中A只能是words中的成员，否则返回-1。要求在调用guess函数不超过10次的条件下，找到谜底。

每次对一个字符串A调用guess函数，如果guess(A)=6，则A就是谜底。否则，利用函数sameChar(A,B)检查剩余字符串B与A的相同位置上相同字符的个数，如果sameChar(A,B)!=guess(A)，B不可能是谜底，淘汰。否则进入下一轮。当然，A也会被淘汰。

但是每轮作为对比标准的字符串A的选择会影响到结果，所以尽量用随机选择的方法。

2.2 代码

class Solution {
public:
    int sameLetter(string&A, string&B){
        int count = 0;
        for(int i = 0; i<A.size(); i++){
            if(A[i]==B[i])
                count+=1;
        }
        return count;
    }
    void findSecretWord(vector<string>& wordlist, Master& master) {
        vector<string> candidates(wordlist.begin(),wordlist.end());
        while(candidates.size()){
            string base = candidates[candidates.size()/2];
            int correct = master.guess(base);
            if(correct == 6) break;
            vector<string> temp;
            for(int i = 0; i < candidates.size();i++){
                if(i != candidates.size()/2 && sameLetter(base,candidates[i])==correct){
                    temp.push_back(candidates[i]);
                }
            }
            candidates.swap(temp);
        }
    }
};

3 标准解

3.1 分析

为了使每次筛选后得到的结果尽可能小，我们每次调用函数前，统计每一对字符串i，j的共同字符数H(i,j)。然后遍历所有候选字符串，统计以每一个字符串i为对比标准，得到与i有0,1,2,3,4,5个相同字符的各组字符串的个数，取其中的最大值max[i]。然后取所有字符串中最小基准min(max[i] for i in cadidates)，即可得到局部最大筛选结果。