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)273Please respect copyright.ＰＥＮＡＮＡIqJPAbaoev
// better than use DFS as it just need to find out the shortest path.
class Solution {273Please respect copyright.ＰＥＮＡＮＡtw6TvsrcvL
public int minMutation(String start, String end, String[] bank) {273Please respect copyright.ＰＥＮＡＮＡaonqBrcRdk
// 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.273Please respect copyright.ＰＥＮＡＮＡzlqHpjzl1H
Queue<String> queue = new LinkedList<>();273Please respect copyright.ＰＥＮＡＮＡCbKSQ5M2WA
Set<String> seen = new HashSet<>();273Please respect copyright.ＰＥＮＡＮＡvbSYkI3EwK
queue.add(start);273Please respect copyright.ＰＥＮＡＮＡmVOiatffch
seen.add(start);273Please respect copyright.ＰＥＮＡＮＡZ6uRIsCmfF
273Please respect copyright.ＰＥＮＡＮＡo6uyDnq4Lb
int steps = 0;273Please respect copyright.ＰＥＮＡＮＡkPIUAPxY7u
273Please respect copyright.ＰＥＮＡＮＡY4vTzqjfdz
while (!queue.isEmpty()) {273Please respect copyright.ＰＥＮＡＮＡaN6PGWrLE5
int nodesInQueue = queue.size();273Please respect copyright.ＰＥＮＡＮＡX0AWT64F2W
for (int j = 0; j < nodesInQueue; j++) {273Please respect copyright.ＰＥＮＡＮＡC10velJio6
String node = queue.remove();273Please respect copyright.ＰＥＮＡＮＡCg9lgSYIju
// 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)) {273Please respect copyright.ＰＥＮＡＮＡzUZy7YK8UX
return steps;273Please respect copyright.ＰＥＮＡＮＡhKuaoQgWsj
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {273Please respect copyright.ＰＥＮＡＮＡECPwgvQLvZ
for (int i = 0; i < node.length(); i++) {273Please respect copyright.ＰＥＮＡＮＡMXO4Ktu9hF
String neighbor = node.substring(0, i) + c + node.substring(i + 1);273Please respect copyright.ＰＥＮＡＮＡwD006nhi23
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {273Please respect copyright.ＰＥＮＡＮＡUQKRoVPFZl
queue.add(neighbor);273Please respect copyright.ＰＥＮＡＮＡzPFq1dZP5l
seen.add(neighbor);273Please respect copyright.ＰＥＮＡＮＡZlJ73TRGUu
}273Please respect copyright.ＰＥＮＡＮＡitlV53sKQH
}273Please respect copyright.ＰＥＮＡＮＡlPA77BCcjH
}273Please respect copyright.ＰＥＮＡＮＡ9RYCL9JqCS
}273Please respect copyright.ＰＥＮＡＮＡv6FZyVId0p
273Please respect copyright.ＰＥＮＡＮＡMDtCPJZ7fy
steps++;273Please respect copyright.ＰＥＮＡＮＡMlw8mj7wdZ
}273Please respect copyright.ＰＥＮＡＮＡ6xLZXoGQkq
// If we finish the BFS and did not find end, return -1.273Please respect copyright.ＰＥＮＡＮＡe5VSAP5Wo0
return -1;273Please respect copyright.ＰＥＮＡＮＡR2B5GeGxqJ
}273Please respect copyright.ＰＥＮＡＮＡIu2Cc6s9CV
}
然後按「加至主畫面」