Word Ladder leetcode

最新推荐文章于 2019-07-03 13:56:36 发布

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最新推荐文章于 2019-07-03 13:56:36 发布

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本文链接：https://blog.youkuaiyun.com/nutcc/article/details/10496315

本文介绍了一种使用广度优先搜索解决单词阶梯问题的方法。给定两个单词和一个字典，算法寻找从起始单词到目标单词的最短转换路径，每次只改变一个字母，并确保中间状态存在于字典中。

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Word Ladder

Feb 11

10499 / 37276

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence fromstart to end, 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;
	}
}