[leetcode]126. Word LadderII@Java解题报告

https://leetcode.com/problems/word-ladder-ii/#/description

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWordto endWord, such that:

1. Only one letter can be changed at a time
2. Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log","cog"]

Return

  [    ["hit","hot","dot","dog","cog"],    ["hit","hot","lot","log","cog"]  ]

Note:

• Return an empty list if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.
• You may assume no duplicates in the word list.
• You may assume beginWord and endWord are non-empty and are not the same.

UPDATE (2017/1/20):
The wordList parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

package go.jacob.day727;import java.util.ArrayList;import java.util.HashMap;import java.util.LinkedList;import java.util.List;import java.util.Map;import java.util.Queue;/** * 126. Word Ladder II * 思路：图bfs的应用 * @author Jacob * */public class Demo2 {// res为最终结果List<List<String>> res;Map<String, List<String>> map;public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {res = new ArrayList<List<String>>();if (wordList.size() == 0)return res;// 最小步数初始化为最大值int min = Integer.MAX_VALUE;Queue<String> queue = new LinkedList<String>();queue.add(beginWord);map = new HashMap<String, List<String>>();// ladder用来记录每个string出现的最小序号Map<String, Integer> ladder = new HashMap<String, Integer>();for (String string : wordList)// 初始化为最大值ladder.put(string, Integer.MAX_VALUE);ladder.put(beginWord, 0);wordList.add(endWord);while (!queue.isEmpty()) {String word = queue.poll();int step = ladder.get(word) + 1;// 如果大于最小步数，直接跳出循环if (step > min)break;for (int i = 0; i < word.length(); i++) {StringBuilder sb = new StringBuilder();for (char c = 'a'; c <= 'z'; c++) {sb.setCharAt(i, c);String new_word = sb.toString();if (ladder.containsKey(new_word)) {// 只有第一遍历到的new_word才会加入到queue,map，因为step是非递减的if (step > ladder.get(new_word))continue;else if (step < ladder.get(new_word)) {queue.add(new_word);ladder.put(new_word, step);} else;// map中只存放word的所有前一个元素if (map.containsKey(new_word))map.get(new_word).add(word);else {List<String> list = new LinkedList<String>();list.add(word);map.put(new_word, list);}if (new_word.equals(endWord))min = step;}}}}//只能用LinkedList，不能用ArrayList。原因见backTrace方法LinkedList<String> result = new LinkedList<String>();backTrace(endWord, beginWord, result);return res;}private void backTrace(String word, String start, List<String> list) {if (word.equals(start)) {list.add(0, start);res.add(new ArrayList<String>(list));list.remove(0);return;}//注意：这里的list是LinkedList类型，所以插入index=0的位置，其余元素全部后移//如果是ArrayList，会把0处的元素替换list.add(0, word);if (map.get(word) != null)for (String s : map.get(word))backTrace(s, start, list);list.remove(0);}}