433. Minimum Genetic Mutation

问题描述：

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:

1. Starting point is assumed to be valid, so it might not be included in the bank.
2. If multiple mutations are needed, all mutations during in the sequence must be valid.
3. 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

解题思路：

这道题可以用bfs来解答，与word ladder十分相似。

不过这道题有更多的限制：

　　1.gene从{‘A‘, ‘C‘, ‘G‘, ‘T‘ }中选择。

　　2.每一个mutation的结果必须在bank中能够找到。

我们用queue来存储当前的gene string，并且用另一个queue来存储变化次数。

这两个queue要保持同步。

在最开始可以首先检查end是否在bank中，不在bank中的话，无论如何我们也到不达终点，可以直接返回-1.

在检查当前字符串是否可以加入队列时需要检查：

　　1. 是否已经访问过

　　2. 是否在bank中可以找到

代码:

class Solution {
public:
    int minMutation(string start, string end, vector<string>& bank) {
        vector<char> genes = {‘A‘, ‘T‘, ‘C‘, ‘G‘};
        unordered_set<string> s(bank.begin(), bank.end());

        if(s.count(end) == 0) return -1;

        unordered_set<string> visited;
        queue<string> q;
        queue<int> distance_q;
        q.push(start);
        distance_q.push(0);

        while(!q.empty()){
            string cur = q.front();
            q.pop();
            int dis = distance_q.front();
            distance_q.pop();
            for(int i = 0 ; i < 8; i++){
                for(int j = 0; j < 4; j++){
                    string temp = cur;
                    if(genes[j] == temp[i]) continue;
                    temp[i] = genes[j];

                    if(temp == end) return dis+1;
                    if(visited.count(temp) == 0 && s.count(temp) != 0){
                        visited.insert(temp);
                        q.push(temp);
                        distance_q.push(dis+1);
                    }
                }
            }
        }
        return -1;
    }
};

原文地址：https://www.cnblogs.com/yaoyudadudu/p/9393852.html