LeetCode #433. Minimum Genetic Mutation_a gene string can be represented by an 8-character-优快云博客

本文探讨了在一个基因库中，从一个起始基因通过最少的单字符变异到达目标基因的问题。利用广度优先搜索算法，文章详细解释了如何在有效基因集中找到最短的变异路径。

题目描述：

A gene string can be represented by an 8-character long string, with choices from "A", "C", "G", "T".

Suppose we need to investigate about a mutation (mutation from "start" to "end"), where ONE mutation is defined as ONE single character changed in the gene string.

For example, "AACCGGTT" -> "AACCGGTA" is 1 mutation.

Also, there is a given gene "bank", which records all the valid gene mutations. A gene must be in the bank to make it a valid gene string.

Now, given 3 things - start, end, bank, your task is to determine what is the minimum number of mutations needed to mutate from "start" to "end". If there is no such a mutation, return -1.

Note:

Starting point is assumed to be valid, so it might not be included in the bank.
If multiple mutations are needed, all mutations during in the sequence must be valid.
You may assume start and end string is not the same.

Example 1:

start: "AACCGGTT"
end:   "AACCGGTA"
bank: ["AACCGGTA"]
return: 1

Example 2:

start: "AACCGGTT"
end:   "AAACGGTA"
bank: ["AACCGGTA", "AACCGCTA", "AAACGGTA"]
return: 2

Example 3:

start: "AAAAACCC"
end:   "AACCCCCC"
bank: ["AAAACCCC", "AAACCCCC", "AACCCCCC"]
return: 3

class Solution {
public:
    int minMutation(string start, string end, vector<string>& bank) {
        unordered_set<string> s;
        for(string& gene:bank) s.insert(gene);
        if(s.count(end)==0) return -1;
        queue<string> q;
        q.push(start);
        s.erase(start);
        unordered_map<string,int> dist;
        dist[start]=0;
        vector<int> chars={'A','C','G','T'};
        while(!q.empty())
        {
            string cur=q.front();
            q.pop();
            if(cur==end) return dist[end];
            for(int i=0;i<cur.size();i++)
            {
                for(char c:chars)
                {
                    string next=cur.substr(0,i)+c+cur.substr(i+1);
                    if(s.count(next))
                    {
                        dist[next]=dist[cur]+1;
                        q.push(next);
                        s.erase(next);
                    }
                }
            }
        }
        return -1;
    }
};