Q: A gene string can be represented by an 8-character long string, with choices from 'A', 'C', 'G', and 'T'.
Suppose we need to investigate a mutation from a gene string start to a gene string end where one mutation is defined as one single character changed in the gene string.
- For example, "AACCGGTT" --> "AACCGGTA" is one mutation.
There is also a gene bank bank that records all the valid gene mutations. A gene must be in bank to make it a valid gene string.
Given the two gene strings start and end and the gene bank bank, return the minimum number of mutations needed to mutate from start to end. If there is no such a mutation, return -1.
Note that the starting point is assumed to be valid, so it might not be included in the bank.
A: BFS (Breadth-First Search)254Please respect copyright.ＰＥＮＡＮＡLqZpqKjznD
// better than use DFS as it just need to find out the shortest path.
class Solution {254Please respect copyright.ＰＥＮＡＮＡR7F1GhIJYc
public int minMutation(String start, String end, String[] bank) {254Please respect copyright.ＰＥＮＡＮＡqkewlmxxS9
// Initialize a queue queue and a set seen. The queue will be used for BFS and the set will be used to prevent visiting a node more than once. Initially, the queue and set should hold start.254Please respect copyright.ＰＥＮＡＮＡRgf8J9YHYH
Queue<String> queue = new LinkedList<>();254Please respect copyright.ＰＥＮＡＮＡtkUgKSi4jR
Set<String> seen = new HashSet<>();254Please respect copyright.ＰＥＮＡＮＡk9SEJjPkBx
queue.add(start);254Please respect copyright.ＰＥＮＡＮＡySiaZm9UXg
seen.add(start);254Please respect copyright.ＰＥＮＡＮＡ7uaQvZ8jI0
254Please respect copyright.ＰＥＮＡＮＡmcha6IWbMj
int steps = 0;254Please respect copyright.ＰＥＮＡＮＡ2n6oflfUeK
254Please respect copyright.ＰＥＮＡＮＡ1qmCEikAAa
while (!queue.isEmpty()) {254Please respect copyright.ＰＥＮＡＮＡePQdpHsmlR
int nodesInQueue = queue.size();254Please respect copyright.ＰＥＮＡＮＡsovIFzGicV
for (int j = 0; j < nodesInQueue; j++) {254Please respect copyright.ＰＥＮＡＮＡ7jWV5WHPN7
String node = queue.remove();254Please respect copyright.ＰＥＮＡＮＡZTIhSfWCbe
// Perform a BFS. At each node, if node == end, return the number of steps so far. Otherwise, iterate over all the neighbors. For each neighbor, if neighbor is not in seen and neighbor is in bank, add it to queue and seen.
if (node.equals(end)) {254Please respect copyright.ＰＥＮＡＮＡKHZkMlKsOr
return steps;254Please respect copyright.ＰＥＮＡＮＡN2unqMZeeC
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {254Please respect copyright.ＰＥＮＡＮＡ2MH5VLdMK9
for (int i = 0; i < node.length(); i++) {254Please respect copyright.ＰＥＮＡＮＡWNOcJ1zaIj
String neighbor = node.substring(0, i) + c + node.substring(i + 1);254Please respect copyright.ＰＥＮＡＮＡ7hv6FHdKSu
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {254Please respect copyright.ＰＥＮＡＮＡthxSPHHXPK
queue.add(neighbor);254Please respect copyright.ＰＥＮＡＮＡT8AGnz9HnG
seen.add(neighbor);254Please respect copyright.ＰＥＮＡＮＡfwOUIbhxM9
}254Please respect copyright.ＰＥＮＡＮＡjQhXhxY59c
}254Please respect copyright.ＰＥＮＡＮＡOYiz03mmtj
}254Please respect copyright.ＰＥＮＡＮＡ9kEzUZPstY
}254Please respect copyright.ＰＥＮＡＮＡa38kS48wCc
254Please respect copyright.ＰＥＮＡＮＡ8CrUYGZx1e
steps++;254Please respect copyright.ＰＥＮＡＮＡjNQFEoxFsr
}254Please respect copyright.ＰＥＮＡＮＡzqmQ1vCUqt
// If we finish the BFS and did not find end, return -1.254Please respect copyright.ＰＥＮＡＮＡ3j7CeAm6mk
return -1;254Please respect copyright.ＰＥＮＡＮＡ6zetWzm0LA
}254Please respect copyright.ＰＥＮＡＮＡpLdf4QL6Ei
}
然後按「加至主畫面」