[LeetCode] Minimum Genetic Mutation 最小基因变化

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

这道题跟之前的 Word Ladder 完全是一道题啊，换个故事就直接来啊，越来不走心了啊。不过博主做的时候并没有想起来是之前一样的题，而是先按照脑海里第一个浮现出的思路做的，发现也通过OJ了。博主使用的一种BFS的搜索，先建立bank数组的距离场，这里距离就是两个字符串之间不同字符的个数。然后以start字符串为起点，向周围距离为1的点扩散，采用BFS搜索，每扩散一层，level自加1，当扩散到end字符串时，返回当前level即可。注意我们要把start字符串也加入bank中，而且此时我们也知道start的坐标位置，bank的最后一个位置，然后在建立距离场的时候，调用一个count子函数，用来统计输入的两个字符串之间不同字符的个数，注意dist[i][j]和dist[j][i]是相同，所以我们只用算一次就行了。然后我们进行BFS搜索，用一个visited集合来保存遍历过的字符串，注意检测距离的时候，dist[i][j]和dist[j][i]只要有一个是1，就可以了，参见代码如下：

解法一：

class Solution {
public:
    int minMutation(string start, string end, vector<string>& bank) {
        if (bank.empty()) return -1;
        bank.push_back(start);
        int res = 0, n = bank.size();
        queue<int> q{{n - 1}};
        unordered_set<int> visited;
        vector<vector<int>> dist(n, vector<int>(n, 0));
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                dist[i][j] = count(bank[i], bank[j]);
            }
        }
        while (!q.empty()) {
            ++res;
            for (int i = q.size(); i > 0; --i) {
                int t = q.front(); q.pop();
                visited.insert(t);
                for (int j = 0; j < n; ++j) {
                    if ((dist[t][j] != 1 && dist[j][t] != 1) || visited.count(j)) continue;
                    if (bank[j] == end) return res;
                    q.push(j);
                }
            }
        }
        return -1;
    }
    int count(string word1, string word2) {
        int cnt = 0, n = word1.size();
        for (int i = 0; i < n; ++i) {
            if (word1[i] != word2[i]) ++cnt;
        }
        return cnt;
    }
};

下面这种解法跟之前的那道 Word Ladder 是一样的，也是用的BFS搜索。跟上面的解法不同之处在于，对于遍历到的字符串，我们不再有距离场，而是对于每个字符，我们都尝试将其换为一个新的字符，每次只换一个，这样会得到一个新的字符串，如果这个字符串在bank中存在，说明这样变换是合法的，加入visited集合和queue中等待下一次遍历，记得在下次置换字符的时候要将之前的还原。我们在queue中取字符串出来遍历的时候，先检测其是否和end相等，相等的话返回level，参见代码如下：

解法二：

class Solution {
public:
    int minMutation(string start, string end, vector<string>& bank) {
        if (bank.empty()) return -1;
        vector<char> gens{'A','C','G','T'};
        unordered_set<string> s{bank.begin(), bank.end()};
        unordered_set<string> visited;
        queue<string> q{{start}};
        int level = 0;
        while (!q.empty()) {
            for (int i = q.size(); i > 0; --i) {
                string t = q.front(); q.pop();
                if (t == end) return level;
                for (int j = 0; j < t.size(); ++j) {
                    char old = t[j];
                    for (char c : gens) {
                        t[j] = c;
                        if (s.count(t) && !visited.count(t)) {
                            visited.insert(t);
                            q.push(t);
                        }
                    }
                    t[j] = old;
                }
            }
            ++level;
        }
        return -1;
    }
};

再来看一种递归的解法，跟 Permutations 中的解法一有些类似，是遍历bank中的字符串，跟当前的字符串cur相比较，调用isDiffOne()函数判断，若正好跟cur相差一个字符，并且之前没有访问过，那么先在visited数组中标记为true，然后调用递归函数，若返回的不为-1，则用其更新结果res，因为-1代表无法变换成cur。调用完递归后恢复状态，在visited数组中标记为false。循环结束后，看res的值，若还是n+1，表示无法更新，返回-1，否则返回res+1，因为这里的res是变换了一次后到达目标字符串的最小变化次数，所以要加上当前的这次变换。至于isDiffOne()函数就没啥难度了，就是一个一个的比较，不同就累加计数器cnt，参见代码如下：

解法三：

class Solution {
public:
    int minMutation(string start, string end, vector<string>& bank) {
        if (bank.empty()) return -1;
        vector<bool> visited(bank.size(), false);
        return helper(start, end, bank, visited);
    }
    int helper(string cur, string end, vector<string>& bank, vector<bool>& visited) {
        if (cur == end) return 0;
        int n = bank.size(), res = n + 1;
        for (int i = 0; i < n; ++i) {
            if (visited[i] || !isDiffOne(bank[i], cur)) continue;
            visited[i] = true;
            int t = helper(bank[i], end, bank, visited);
            if (t != -1) res = min(res, t);
            visited[i] = false;
        }
        return res == n + 1 ? -1 : res + 1;
    }
    bool isDiffOne(string& s1, string& s2) {
        int cnt = 0, n = s1.size();
        for (int i = 0; i < n; ++i) {
            if (s1[i] != s2[i]) ++cnt;
            if (cnt > 1) break;
        }
        return cnt == 1;
    }
};