x
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)139Please respect copyright.ＰＥＮＡＮＡ7sWHhi0kHf
// better than use DFS as it just need to find out the shortest path.
class Solution {139Please respect copyright.ＰＥＮＡＮＡ8BOr3fWP99
public int minMutation(String start, String end, String[] bank) {139Please respect copyright.ＰＥＮＡＮＡNc2fy5WPRi
// 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.139Please respect copyright.ＰＥＮＡＮＡ4G30Kwv1hG
Queue<String> queue = new LinkedList<>();139Please respect copyright.ＰＥＮＡＮＡlFahuG9UUa
Set<String> seen = new HashSet<>();139Please respect copyright.ＰＥＮＡＮＡS3STkgqCOK
queue.add(start);139Please respect copyright.ＰＥＮＡＮＡhG6lKlR5cM
seen.add(start);139Please respect copyright.ＰＥＮＡＮＡoOnG7GgNWK
139Please respect copyright.ＰＥＮＡＮＡ0NORj9tQ4a
int steps = 0;139Please respect copyright.ＰＥＮＡＮＡbgV18abgzX
139Please respect copyright.ＰＥＮＡＮＡjGUN4IzNJB
while (!queue.isEmpty()) {139Please respect copyright.ＰＥＮＡＮＡWIQKzsrxeg
int nodesInQueue = queue.size();139Please respect copyright.ＰＥＮＡＮＡmcAHkjdKg0
for (int j = 0; j < nodesInQueue; j++) {139Please respect copyright.ＰＥＮＡＮＡKBUpdJnGty
String node = queue.remove();139Please respect copyright.ＰＥＮＡＮＡSUm3tGYJ1O
// 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)) {139Please respect copyright.ＰＥＮＡＮＡ1hCra56RR5
return steps;139Please respect copyright.ＰＥＮＡＮＡiH6aPqOLrt
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {139Please respect copyright.ＰＥＮＡＮＡXXKMRw1sDT
for (int i = 0; i < node.length(); i++) {139Please respect copyright.ＰＥＮＡＮＡrOk2rDJApf
String neighbor = node.substring(0, i) + c + node.substring(i + 1);139Please respect copyright.ＰＥＮＡＮＡhs5Zo68Yci
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {139Please respect copyright.ＰＥＮＡＮＡGkaAOILmSC
queue.add(neighbor);139Please respect copyright.ＰＥＮＡＮＡJ5pLejX4Eq
seen.add(neighbor);139Please respect copyright.ＰＥＮＡＮＡtMN4big9wr
}139Please respect copyright.ＰＥＮＡＮＡEZbVERub2m
}139Please respect copyright.ＰＥＮＡＮＡqNYJSQxBVU
}139Please respect copyright.ＰＥＮＡＮＡBBPYZmzet7
}139Please respect copyright.ＰＥＮＡＮＡP2Htq3uWiY
139Please respect copyright.ＰＥＮＡＮＡUStnjf76KG
steps++;139Please respect copyright.ＰＥＮＡＮＡgm9SGrCAOa
}139Please respect copyright.ＰＥＮＡＮＡ9x1AiMspbw
// If we finish the BFS and did not find end, return -1.139Please respect copyright.ＰＥＮＡＮＡW60aweXrMi
return -1;139Please respect copyright.ＰＥＮＡＮＡvCXRITxz6e
}139Please respect copyright.ＰＥＮＡＮＡ36cuf8GcOO
}
