127. Word Ladder

forsigner 发布于2019-08-14 14:25 / 655人阅读

摘要：题目解答主要解题思路的，把每一种可能的都放进去试，看能不能有一条线边到代码当然，这样的时间还不是最优化的，如果我们从两头扫，扫到中间任何一个能够串联起来都可以，如果没有找到可以串联的那么返回。

题目：
Given two words (beginWord and endWord), and a dictionary"s word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

Only one letter can be changed at a time
Each intermediate word must exist in the word list
For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["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.

解答：
主要解题思路的bfs，把每一种可能的character都放进去试，看能不能有一条线边到endWord.
代码：

public class Solution {
    public int ladderLength(String beginWord, String endWord, Set wordList) {
        //BFS to solve the problem
        int count = 1;
        Set reached = new HashSet();
        reached.add(beginWord);
        wordList.add(endWord);
        
        while (!reached.contains(endWord)) {
            Set toAdd = new HashSet();
            for (String word : reached) {
                
                for (int i = 0; i < word.length(); i++) {
                    char[] chars = word.toCharArray();
                    for (char c = "a"; c <= "z"; c++) {
                        chars[i] = c;
                        String newWord = String.valueOf(chars);
                        if (wordList.contains(newWord)) {
                            toAdd.add(newWord);
                            wordList.remove(newWord);
                        }
                    }
                }
            }
            count++;
            if (toAdd.size() == 0) return 0;
            reached = toAdd;
        }
        return count;
    }
}

当然，这样的时间还不是最优化的，如果我们从两头扫，扫到中间任何一个word能够串联起来都可以，如果没有找到可以串联的word,那么返回0。代码如下：

public class Solution {
    public int ladderLength(String beginWord, String endWord, Set wordList) {
        int count = 1;
        Set beginSet = new HashSet();
        Set endSet = new HashSet();
        Set visited = new HashSet();
        beginSet.add(beginWord);
        endSet.add(endWord);
        
        while (!beginSet.isEmpty() && !endSet.isEmpty()) {
            
            
            if (beginSet.size() > endSet.size()) {
                Set temp = beginSet;
                beginSet = endSet;
                endSet = temp;
            }
            
            Set toAdd = new HashSet();
            for (String word : beginSet) {
                for (int i = 0; i < word.length(); i++) {
                    char[] chars = word.toCharArray();
                    for (char c = "a"; c <= "z"; c++) {
                        chars[i] = c;
                        String newWord = String.valueOf(chars);
                        if (endSet.contains(newWord)) return count + 1;
                        if (!visited.contains(newWord) && wordList.contains(newWord)) {
                            toAdd.add(newWord);
                            visited.add(newWord);
                        }
                    }
                }
            }
            count++;
            beginSet = toAdd;
        }
        return 0;
    }
}

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