https://oj.leetcode.com/problems/word-ladder/
http://blog.csdn.net/linhuanmars/article/category/1918893/2
public class Solution {
public int ladderLength(String start, String end, Set<String> dict)
{
// Put end into dict
Set<String> dictionary = new HashSet<>(dict);
dictionary.add(end);
Set<String> visited = new HashSet<>();
// BFS
int nextlevel = 0;
int curlevel = 1;
int result = 1;
Queue<String> queue = new LinkedList<>();
queue.offer(start);
visited.add(start);
while (!queue.isEmpty())
{
String node = queue.poll();
if (node.equals(end))
return result;
curlevel --;
Set<String> nextstrs = findnext(node, dictionary, visited);
for (String nextstr : nextstrs)
{
queue.offer(nextstr);
}
nextlevel += nextstrs.size();
if (curlevel == 0)
{
// This level is finished
curlevel = nextlevel;
nextlevel = 0;
result ++;
}
}
return 0;
}
// Given string, find next str from dict (excluding visited)
private Set<String> findnext(String from, Set<String> dict, Set<String> visited)
{
Set<String> toReturn = new HashSet<>();
char[] chars = from.toCharArray();
for (int i = 0 ; i < chars.length ; i ++)
{
char originalc = chars[i];
for (char j = 'a' ; j <= 'z' ; j ++)
{
if (originalc == j)
continue;
chars[i] = j;
String str = new String(chars);
if (dict.contains(str) && !visited.contains(str))
{
toReturn.add(str);
// NOTE
// Pre put str into visited, because we will anyway visited
// This is to increase latency.
visited.add(str);
}
}
// NOTE
// Don't forget change it back.
chars[i] = originalc;
}
return toReturn;
}
}
