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)234Please respect copyright.ＰＥＮＡＮＡ5VMcvRxXJy
// better than use DFS as it just need to find out the shortest path.
class Solution {234Please respect copyright.ＰＥＮＡＮＡay2cwhE7uv
public int minMutation(String start, String end, String[] bank) {234Please respect copyright.ＰＥＮＡＮＡJmldHPMfyv
// 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.234Please respect copyright.ＰＥＮＡＮＡye7qOTLsPj
Queue<String> queue = new LinkedList<>();234Please respect copyright.ＰＥＮＡＮＡITRXbLQXGZ
Set<String> seen = new HashSet<>();234Please respect copyright.ＰＥＮＡＮＡcWf9PvahFs
queue.add(start);234Please respect copyright.ＰＥＮＡＮＡFHEMwcFwFO
seen.add(start);234Please respect copyright.ＰＥＮＡＮＡn9kbA12Hfi
234Please respect copyright.ＰＥＮＡＮＡlVoRWNuwrU
int steps = 0;234Please respect copyright.ＰＥＮＡＮＡOVHwydLBMw
234Please respect copyright.ＰＥＮＡＮＡMIsIafQwFO
while (!queue.isEmpty()) {234Please respect copyright.ＰＥＮＡＮＡszFtCQAWtJ
int nodesInQueue = queue.size();234Please respect copyright.ＰＥＮＡＮＡm51jQaccjE
for (int j = 0; j < nodesInQueue; j++) {234Please respect copyright.ＰＥＮＡＮＡtaNdoFEgp3
String node = queue.remove();234Please respect copyright.ＰＥＮＡＮＡzWAfb3ksHq
// 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)) {234Please respect copyright.ＰＥＮＡＮＡIzmaRI62Q7
return steps;234Please respect copyright.ＰＥＮＡＮＡjUjhVYoCBF
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {234Please respect copyright.ＰＥＮＡＮＡMKw0a5Juvx
for (int i = 0; i < node.length(); i++) {234Please respect copyright.ＰＥＮＡＮＡhCVccatYnw
String neighbor = node.substring(0, i) + c + node.substring(i + 1);234Please respect copyright.ＰＥＮＡＮＡ8HnmQtMMVj
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {234Please respect copyright.ＰＥＮＡＮＡWXyQTWqhdh
queue.add(neighbor);234Please respect copyright.ＰＥＮＡＮＡtJ3LeRjIq0
seen.add(neighbor);234Please respect copyright.ＰＥＮＡＮＡw3mp7nF7Kb
}234Please respect copyright.ＰＥＮＡＮＡmruLMaEmUc
}234Please respect copyright.ＰＥＮＡＮＡkIApAX6fW2
}234Please respect copyright.ＰＥＮＡＮＡhgOb6UaXnn
}234Please respect copyright.ＰＥＮＡＮＡW0iyWy0oEM
234Please respect copyright.ＰＥＮＡＮＡqFG1y0pAM8
steps++;234Please respect copyright.ＰＥＮＡＮＡICXTgr74FW
}234Please respect copyright.ＰＥＮＡＮＡgRulEnUhHa
// If we finish the BFS and did not find end, return -1.234Please respect copyright.ＰＥＮＡＮＡtjVNV0tpYe
return -1;234Please respect copyright.ＰＥＮＡＮＡTHC6CyC87D
}234Please respect copyright.ＰＥＮＡＮＡiBT0asf7xb
}
然後按「加至主畫面」