Word Ladder leetcode

Word Ladder

Feb 11

10499 / 37276

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

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

For 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

public class Solution {     public int ladderLength(String start, String end, HashSet<String> dict) {Queue<String> queue = new LinkedList<String>();queue.offer(start);HashSet<String> used = new HashSet<String>();int step = 1;int l1 = 1,l2 = 0;  //很重要！ 用来记录当前的层数。。while (queue.size() != 0) {String s = queue.poll(); ////System.out.println("now=" + s);used.add(s); // 忘记放到used里面了...//System.out.println(used.size());HashSet<String> strs = getNext(s,end, dict, used);l2+=strs.size();l1--;for (String ss : strs) {if (ss.equals(end)) {return step + 1;} else {queue.offer(ss); //写错了》。。。}}if(l1==0){step++;l1 = l2;l2 = 0;}}return 0;}public HashSet<String> getNext(String s, String end,HashSet<String> dict,HashSet<String> used) {if (s == null)return null;HashSet<String> set = new HashSet<String>();for (int i = 0; i < s.length(); i++) {StringBuilder now = new StringBuilder(s); //for (char c = 'a'; c <= 'z'; c++) {now.setCharAt(i, c); //// System.out.println("@@"+now.toString());if (end.equals(now.toString())||dict.contains(now.toString())&& !used.contains(now.toString())) { // toString!!!!!!!!!set.add(now.toString()); ////System.out.println("next" + now.toString());}}}return set;}}