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)200Please respect copyright.ＰＥＮＡＮＡV6lgrnvvig
// better than use DFS as it just need to find out the shortest path.
class Solution {200Please respect copyright.ＰＥＮＡＮＡBB8My2xZCA
public int minMutation(String start, String end, String[] bank) {200Please respect copyright.ＰＥＮＡＮＡJLDmr75N1A
// 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.200Please respect copyright.ＰＥＮＡＮＡ3bgIEJQ2U4
Queue<String> queue = new LinkedList<>();200Please respect copyright.ＰＥＮＡＮＡpxbegh8PED
Set<String> seen = new HashSet<>();200Please respect copyright.ＰＥＮＡＮＡgYYWNHR6Mt
queue.add(start);200Please respect copyright.ＰＥＮＡＮＡNvHxOcWugH
seen.add(start);200Please respect copyright.ＰＥＮＡＮＡp4mL7ZSpab
200Please respect copyright.ＰＥＮＡＮＡM0bSYyNck8
int steps = 0;200Please respect copyright.ＰＥＮＡＮＡ9ruzlQBbtv
200Please respect copyright.ＰＥＮＡＮＡ7zDuIOFZnO
while (!queue.isEmpty()) {200Please respect copyright.ＰＥＮＡＮＡ1XVMGtx4O9
int nodesInQueue = queue.size();200Please respect copyright.ＰＥＮＡＮＡ7q9TmSidyx
for (int j = 0; j < nodesInQueue; j++) {200Please respect copyright.ＰＥＮＡＮＡo0jBM8Nazf
String node = queue.remove();200Please respect copyright.ＰＥＮＡＮＡcf0ci5jHtK
// 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)) {200Please respect copyright.ＰＥＮＡＮＡuDYEechVpv
return steps;200Please respect copyright.ＰＥＮＡＮＡoMGS5I6fe5
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {200Please respect copyright.ＰＥＮＡＮＡaWRTWLbGgo
for (int i = 0; i < node.length(); i++) {200Please respect copyright.ＰＥＮＡＮＡ6vNv5LEmSX
String neighbor = node.substring(0, i) + c + node.substring(i + 1);200Please respect copyright.ＰＥＮＡＮＡTQ7nqIpMta
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {200Please respect copyright.ＰＥＮＡＮＡLu37Ou9OYS
queue.add(neighbor);200Please respect copyright.ＰＥＮＡＮＡVodaSyH4D8
seen.add(neighbor);200Please respect copyright.ＰＥＮＡＮＡT5IrCYq6H1
}200Please respect copyright.ＰＥＮＡＮＡuHiEbsoUHV
}200Please respect copyright.ＰＥＮＡＮＡBTZy49occO
}200Please respect copyright.ＰＥＮＡＮＡT1SqR4s0VI
}200Please respect copyright.ＰＥＮＡＮＡZjrhfPC5rZ
200Please respect copyright.ＰＥＮＡＮＡuSVG7MnxPa
steps++;200Please respect copyright.ＰＥＮＡＮＡoACbYwusZ0
}200Please respect copyright.ＰＥＮＡＮＡtykgZ0XhYj
// If we finish the BFS and did not find end, return -1.200Please respect copyright.ＰＥＮＡＮＡjzG2h8O6fn
return -1;200Please respect copyright.ＰＥＮＡＮＡnFDCsNXAPC
}200Please respect copyright.ＰＥＮＡＮＡleW1ekzO1v
}
