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)197Please respect copyright.ＰＥＮＡＮＡvX0PootC8u
// better than use DFS as it just need to find out the shortest path.
class Solution {197Please respect copyright.ＰＥＮＡＮＡkDp898OnYW
public int minMutation(String start, String end, String[] bank) {197Please respect copyright.ＰＥＮＡＮＡnfmqPeZNhu
// 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.197Please respect copyright.ＰＥＮＡＮＡ63WxDEd5Eh
Queue<String> queue = new LinkedList<>();197Please respect copyright.ＰＥＮＡＮＡPUnfeIRDhm
Set<String> seen = new HashSet<>();197Please respect copyright.ＰＥＮＡＮＡStUqx8ItkO
queue.add(start);197Please respect copyright.ＰＥＮＡＮＡwDfk5oqpjW
seen.add(start);197Please respect copyright.ＰＥＮＡＮＡPamEWkJkGL
197Please respect copyright.ＰＥＮＡＮＡN9cBSO1bcJ
int steps = 0;197Please respect copyright.ＰＥＮＡＮＡt6gP6n5ufo
197Please respect copyright.ＰＥＮＡＮＡFhKlCo6Q0i
while (!queue.isEmpty()) {197Please respect copyright.ＰＥＮＡＮＡdsvaw3JPOS
int nodesInQueue = queue.size();197Please respect copyright.ＰＥＮＡＮＡXnT0rTixH7
for (int j = 0; j < nodesInQueue; j++) {197Please respect copyright.ＰＥＮＡＮＡ9sW7GI4PST
String node = queue.remove();197Please respect copyright.ＰＥＮＡＮＡSPh02jZmCp
// 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)) {197Please respect copyright.ＰＥＮＡＮＡq3rZAewJDi
return steps;197Please respect copyright.ＰＥＮＡＮＡU7vM1ZrUd9
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {197Please respect copyright.ＰＥＮＡＮＡeYACDj6P7x
for (int i = 0; i < node.length(); i++) {197Please respect copyright.ＰＥＮＡＮＡ4noHFZoIP4
String neighbor = node.substring(0, i) + c + node.substring(i + 1);197Please respect copyright.ＰＥＮＡＮＡWrg6e0sIsL
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {197Please respect copyright.ＰＥＮＡＮＡtQVXv78xWm
queue.add(neighbor);197Please respect copyright.ＰＥＮＡＮＡqTpsrTmkP7
seen.add(neighbor);197Please respect copyright.ＰＥＮＡＮＡE0iaUupkiI
}197Please respect copyright.ＰＥＮＡＮＡ1csjVQlfwW
}197Please respect copyright.ＰＥＮＡＮＡT42BQSVThg
}197Please respect copyright.ＰＥＮＡＮＡcpwYDOaAkC
}197Please respect copyright.ＰＥＮＡＮＡR9GF8Ffm2r
197Please respect copyright.ＰＥＮＡＮＡh7NznypEzn
steps++;197Please respect copyright.ＰＥＮＡＮＡcNQlOxkv7E
}197Please respect copyright.ＰＥＮＡＮＡ36CoH6wloW
// If we finish the BFS and did not find end, return -1.197Please respect copyright.ＰＥＮＡＮＡHxBcxorZzu
return -1;197Please respect copyright.ＰＥＮＡＮＡJ5LimQYgPH
}197Please respect copyright.ＰＥＮＡＮＡtCvssqPASB
}
然後按「加至主畫面」