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Description
给出两个单词(start和end)和一个字典,找到从start到end的最短转换序列
比如:
- 每次只能改变一个字母。
- 变换过程中的中间单词必须在字典中出现。
注意事项
- 如果没有转换序列则返回0。
- 所有单词具有相同的长度。
- 所有单词都只包含小写字母。
Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:
- Only one letter can be changed at a time
- Each intermediate word must exist in the dictionary
Notice
- Return 0 if there is no such transformation sequence.
- All words have the same length.
- All words contain only lowercase alphabetic characters.
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
.
Solution
思路:
1. 使用BFS分层搜索每层可能的下一个单词,并把层号加一。
2. 遍历下个单词的所有可能,如果已出现过这个可能则直接跳过,将没出现过的记录到Set和Queue中。
3. 如果下个单词为end,则返回层数。如果知道队列为空都没找到,则返回0。
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) {
// write your code here
if (dict == null) {
return 0;
}
if (start.equals(end)) {
return 1;
}
//为了能搜出这个结果,end需在字典中
dict.add(start);
dict.add(end);
//BFS搜索每层可能的结果,并比较知否找到end
Queue<String> queue = new LinkedList<>();
HashSet<String> set = new HashSet<>();
queue.offer(start);
set.add(start);
int length = 1;
while (!queue.isEmpty()) {
length++; //遍历每层可能的结果前计数值加一
int size = queue.size();
for (int i = 0; i < size; i++) {
String word = queue.poll();
//遍历下个单词的所有可能,如果已出现过这个可能则直接跳过,将没出现过的记录下来
for (String nextWord : getNextWords(word, dict)) {
if (set.contains(nextWord)) {
continue;
}
if (nextWord.equals(end)) {
return length;
}
set.add(nextWord);
queue.offer(nextWord);
}
}
}
return 0;
}
//找到所有可能的结果,看看是不是在字典中有出现,如果出现就记录下来
// get connections with given word.
// for example, given word = 'hot', dict = {'hot', 'hit', 'hog'}
// it will return ['hit', 'hog']
private ArrayList<String> getNextWords(String word, Set<String> dict) {
ArrayList<String> nextWords = new ArrayList<>();
for (char c = 'a'; c < 'z'; c++) {
for (int i = 0; i < word.length(); i++) {
if (c == word.charAt(i)) {
continue;
}
String nextWord = replace(word, i, c);
if (dict.contains(nextWord)) {
nextWords.add(nextWord);
}
}
}
return nextWords;
}
// replace character of a string at given index to a given character
// return a new string
private String replace(String s, int index, char c) {
char[] chars = s.toCharArray();
chars[index] = c;
return new String(chars);
}
}