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)210Please respect copyright.ＰＥＮＡＮＡ0PdPPMFozg
// better than use DFS as it just need to find out the shortest path.
class Solution {210Please respect copyright.ＰＥＮＡＮＡol5ds3KcUO
public int minMutation(String start, String end, String[] bank) {210Please respect copyright.ＰＥＮＡＮＡkbYK0u42H0
// 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.210Please respect copyright.ＰＥＮＡＮＡtpk60x05o4
Queue<String> queue = new LinkedList<>();210Please respect copyright.ＰＥＮＡＮＡu12risv3MV
Set<String> seen = new HashSet<>();210Please respect copyright.ＰＥＮＡＮＡwFkKsSJspA
queue.add(start);210Please respect copyright.ＰＥＮＡＮＡsHp23eopeh
seen.add(start);210Please respect copyright.ＰＥＮＡＮＡpbDkPtA9HV
210Please respect copyright.ＰＥＮＡＮＡINZ1D4wFbM
int steps = 0;210Please respect copyright.ＰＥＮＡＮＡtiVJ6qY7p9
210Please respect copyright.ＰＥＮＡＮＡDJnSjbCBfw
while (!queue.isEmpty()) {210Please respect copyright.ＰＥＮＡＮＡuS20NC2aIr
int nodesInQueue = queue.size();210Please respect copyright.ＰＥＮＡＮＡt8NnaGa3Xf
for (int j = 0; j < nodesInQueue; j++) {210Please respect copyright.ＰＥＮＡＮＡHiIBSLhbcV
String node = queue.remove();210Please respect copyright.ＰＥＮＡＮＡK20fyA0fyU
// 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)) {210Please respect copyright.ＰＥＮＡＮＡpkny4yMIoX
return steps;210Please respect copyright.ＰＥＮＡＮＡT6LZuTNIZH
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {210Please respect copyright.ＰＥＮＡＮＡBpnKTaBEjK
for (int i = 0; i < node.length(); i++) {210Please respect copyright.ＰＥＮＡＮＡoPlk7HIImc
String neighbor = node.substring(0, i) + c + node.substring(i + 1);210Please respect copyright.ＰＥＮＡＮＡBFJ8xF6BAK
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {210Please respect copyright.ＰＥＮＡＮＡey4IBnQ9RX
queue.add(neighbor);210Please respect copyright.ＰＥＮＡＮＡ1KTlfavIRn
seen.add(neighbor);210Please respect copyright.ＰＥＮＡＮＡCnshdVWqUf
}210Please respect copyright.ＰＥＮＡＮＡaLkC7aOrRX
}210Please respect copyright.ＰＥＮＡＮＡHxw7iu5ynE
}210Please respect copyright.ＰＥＮＡＮＡQjXqhBDqwR
}210Please respect copyright.ＰＥＮＡＮＡeFsr5BxsBl
210Please respect copyright.ＰＥＮＡＮＡFOI5sMtod8
steps++;210Please respect copyright.ＰＥＮＡＮＡCigd8Nmdmy
}210Please respect copyright.ＰＥＮＡＮＡ1RY1YXonV6
// If we finish the BFS and did not find end, return -1.210Please respect copyright.ＰＥＮＡＮＡnoZE4ssQib
return -1;210Please respect copyright.ＰＥＮＡＮＡPWQDzItCw7
}210Please respect copyright.ＰＥＮＡＮＡONqyovaJCD
}
然後按「加至主畫面」