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)261Please respect copyright.ＰＥＮＡＮＡqT4yo9njVQ
// better than use DFS as it just need to find out the shortest path.
class Solution {261Please respect copyright.ＰＥＮＡＮＡHLVoJVWeIJ
public int minMutation(String start, String end, String[] bank) {261Please respect copyright.ＰＥＮＡＮＡcpA4RfwbRP
// 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.261Please respect copyright.ＰＥＮＡＮＡVkBetluIGW
Queue<String> queue = new LinkedList<>();261Please respect copyright.ＰＥＮＡＮＡhGYJc4Cylt
Set<String> seen = new HashSet<>();261Please respect copyright.ＰＥＮＡＮＡVis44xBpAb
queue.add(start);261Please respect copyright.ＰＥＮＡＮＡliH8JNuEKv
seen.add(start);261Please respect copyright.ＰＥＮＡＮＡdOfJXHnOHn
261Please respect copyright.ＰＥＮＡＮＡkcjqxF5sq5
int steps = 0;261Please respect copyright.ＰＥＮＡＮＡm9VY4xpJSL
261Please respect copyright.ＰＥＮＡＮＡ9t5P9FcPkx
while (!queue.isEmpty()) {261Please respect copyright.ＰＥＮＡＮＡPjN2S7T4ge
int nodesInQueue = queue.size();261Please respect copyright.ＰＥＮＡＮＡsuBZhz2P0j
for (int j = 0; j < nodesInQueue; j++) {261Please respect copyright.ＰＥＮＡＮＡ4T2eGDDIeL
String node = queue.remove();261Please respect copyright.ＰＥＮＡＮＡgK6h6Vvve5
// 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)) {261Please respect copyright.ＰＥＮＡＮＡH9M2VWeM4k
return steps;261Please respect copyright.ＰＥＮＡＮＡXc9hrZnnGn
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {261Please respect copyright.ＰＥＮＡＮＡ5blXcqdxR7
for (int i = 0; i < node.length(); i++) {261Please respect copyright.ＰＥＮＡＮＡa4wRqu1oMJ
String neighbor = node.substring(0, i) + c + node.substring(i + 1);261Please respect copyright.ＰＥＮＡＮＡwAb4tqbRBG
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {261Please respect copyright.ＰＥＮＡＮＡ2xOghiRUdi
queue.add(neighbor);261Please respect copyright.ＰＥＮＡＮＡzUw1kDtTTR
seen.add(neighbor);261Please respect copyright.ＰＥＮＡＮＡAP7Y7mN9pu
}261Please respect copyright.ＰＥＮＡＮＡDbNtFLlc53
}261Please respect copyright.ＰＥＮＡＮＡsukfwZMhak
}261Please respect copyright.ＰＥＮＡＮＡ8kzvgGa6EY
}261Please respect copyright.ＰＥＮＡＮＡCfa6coNetP
261Please respect copyright.ＰＥＮＡＮＡSA2dI9zZXO
steps++;261Please respect copyright.ＰＥＮＡＮＡsfdoSMWGqn
}261Please respect copyright.ＰＥＮＡＮＡJoDohhmzBW
// If we finish the BFS and did not find end, return -1.261Please respect copyright.ＰＥＮＡＮＡ53w5Nln8Nv
return -1;261Please respect copyright.ＰＥＮＡＮＡTVeMOFpoAz
}261Please respect copyright.ＰＥＮＡＮＡ80xvdcGR8x
}
然後按「加至主畫面」