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)275Please respect copyright.ＰＥＮＡＮＡ36iHBXwcnE
// better than use DFS as it just need to find out the shortest path.
class Solution {275Please respect copyright.ＰＥＮＡＮＡACSqdwRbOv
public int minMutation(String start, String end, String[] bank) {275Please respect copyright.ＰＥＮＡＮＡvmqz8HXsZq
// 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.275Please respect copyright.ＰＥＮＡＮＡ6nwhU6Et1p
Queue<String> queue = new LinkedList<>();275Please respect copyright.ＰＥＮＡＮＡa1dxueKpzP
Set<String> seen = new HashSet<>();275Please respect copyright.ＰＥＮＡＮＡvLtpQo8rO0
queue.add(start);275Please respect copyright.ＰＥＮＡＮＡrjvOXzEA3Y
seen.add(start);275Please respect copyright.ＰＥＮＡＮＡhmlD2Dm4zx
275Please respect copyright.ＰＥＮＡＮＡX8K4Nes58h
int steps = 0;275Please respect copyright.ＰＥＮＡＮＡci3QhsNxxK
275Please respect copyright.ＰＥＮＡＮＡkp5B3B5Zf3
while (!queue.isEmpty()) {275Please respect copyright.ＰＥＮＡＮＡTAwZ4JyZlb
int nodesInQueue = queue.size();275Please respect copyright.ＰＥＮＡＮＡU2BhPU7nwf
for (int j = 0; j < nodesInQueue; j++) {275Please respect copyright.ＰＥＮＡＮＡpJJSgCmR4a
String node = queue.remove();275Please respect copyright.ＰＥＮＡＮＡt2hoDSmVei
// 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)) {275Please respect copyright.ＰＥＮＡＮＡDAuVRMfCea
return steps;275Please respect copyright.ＰＥＮＡＮＡR8AsWWK9Bt
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {275Please respect copyright.ＰＥＮＡＮＡP8jDmXIoFB
for (int i = 0; i < node.length(); i++) {275Please respect copyright.ＰＥＮＡＮＡA23CuYxJJD
String neighbor = node.substring(0, i) + c + node.substring(i + 1);275Please respect copyright.ＰＥＮＡＮＡddVQZYtISd
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {275Please respect copyright.ＰＥＮＡＮＡmIwjpv48yb
queue.add(neighbor);275Please respect copyright.ＰＥＮＡＮＡ0WVznXtdLw
seen.add(neighbor);275Please respect copyright.ＰＥＮＡＮＡKCjdGUGM6M
}275Please respect copyright.ＰＥＮＡＮＡIm0zgPvI9N
}275Please respect copyright.ＰＥＮＡＮＡd6EQ0rzrQL
}275Please respect copyright.ＰＥＮＡＮＡ1GDMOCZsaE
}275Please respect copyright.ＰＥＮＡＮＡEHshawsfft
275Please respect copyright.ＰＥＮＡＮＡNK5aV7dt4M
steps++;275Please respect copyright.ＰＥＮＡＮＡrXkQyVdr8N
}275Please respect copyright.ＰＥＮＡＮＡPKRnVkAU8R
// If we finish the BFS and did not find end, return -1.275Please respect copyright.ＰＥＮＡＮＡAZTDofDaIq
return -1;275Please respect copyright.ＰＥＮＡＮＡ7z9SHP4YWn
}275Please respect copyright.ＰＥＮＡＮＡ7DDABmbXUH
}
