Word Ladder（LintCode）

Word Ladder

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

1. Only one letter can be changed at a time
2. Each intermediate word must exist in the dictionary

Example

Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]

As one shortest transformation is"hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Note

• Return 0 if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.

用BFS做，字符串只相差一个字符的就当作有边。

public class Solution {
    /**
      * @param start, a string
      * @param end, a string
      * @param dict, a set of string
      * @return an integer
      */
    public int ladderLength(String start, String end, Set<String> dict) {
        Queue q = new LinkedList();
        dict.add(end);
        int reslut = 0;
        
        //转成List方便标记,其实也可以将搜索过的String从Set中移除
        List<String> list = new ArrayList<String>(dict);
        int[] flag = new int[list.size()];
        q.add(start);
        int n = 1;
        int x = 0;
        while(q.size() > 0) {
            String y = (String)q.remove();
            n--;
            for (int i = 0;i<list.size() ;i++ ) {
                if(end.equals(y)) return reslut+1;
                if (judge(list.get(i),y) == 1 && flag[i] == 0) {
                    x++;
                    q.add(list.get(i));
                    flag[i] = 1;
                }
            }
            if (n == 0) {
                n = x;
                x = 0;
                reslut++;
            }
        }

        return 0;
    }

    public int judge(String a,String b) {
        int num = 0;
        char[] acs = a.toCharArray();
        char[] bcs = b.toCharArray();
        for (int i = 0;i<acs.length ;i++ ) {
            if(acs[i] != bcs[i]) num++;
        }
        return num;
    }
}