LeetCode :Word Ladder II My Solution

Word Ladder II

 Total Accepted: 11755 Total Submissions: 102776My Submissions

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that:

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

For example,

Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]

Return

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

Note:

• All words have the same length.
• All words contain only lowercase alphabetic characters.

这个题目着实用了不少时间,中间做了两个取巧的处理:

1.把end放到dict里面这样的好处就是在BFS建立图的时候 可以直接获取end的前继,所有.这样可以减少最后处理的麻烦

2.把dfs中处理进行优化,可以直接处理到当前继没有的时候,这个时候最后一个节点一定是start

3.中间建立一个MAP map中保存不同的str对应的前继

4.如果一个串被发现了,那么要看一下是不是distance和之前的str差一个,这个是为了保证不出现绕环

5.对str的每一位做变换,使时间复杂度下来.而刚开始的时候是比较有一个位不相同,但是对海量的处理的时候反而会降低速度.

6.对最后结果要把新的放在0位，oneAnswer.add(0,end).因为是从后往前扫

public class Solution {public ArrayList<ArrayList<String>> findLadders(String start,String end, HashSet<String> dict) {dict.add(end);ArrayList<ArrayList<String>> result = new ArrayList<ArrayList<String>>();if (start == null || end == null || start.length() != end.length()) {return result;}Map<String, MyNode> map = new HashMap<String, MyNode>();Queue<String> queue = new LinkedList<String>();queue.offer(start);map.put(start, new MyNode(start, 1));while (!queue.isEmpty()) {String cur = queue.poll();if (reachEnd(cur, end)) {outputResult(end, new ArrayList<String>(), map, result);return result;}for (int i = 0; i < cur.length(); i++) {for (char c = 'a'; c <= 'z'; c++) {String changeStr = getOneDiff(cur, i, c);if (dict.contains(changeStr)) {MyNode curNode = map.get(cur);int curDist = curNode.distance;int newDist = curDist + 1;if (!map.containsKey(changeStr)) {MyNode newNode = new MyNode(changeStr, newDist);newNode.pre.add(curNode);queue.offer(changeStr);map.put(changeStr, newNode);} else {MyNode preNode = map.get(changeStr);int preDist = preNode.distance;if (newDist == preDist) {preNode.pre.add(curNode);}}}}}}return result;}private void outputResult(String end, ArrayList<String> oneAnswer,Map<String, MyNode> map, ArrayList<ArrayList<String>> result) {MyNode curNode = map.get(end);oneAnswer.add(0, end);if (curNode.pre.isEmpty()) {result.add(new ArrayList<String>(oneAnswer));return;}for (MyNode eachNode : curNode.pre) {outputResult(eachNode.val, oneAnswer, map, result);oneAnswer.remove(0);}}private void getPaths(MyNode myNode, Map<String, MyNode> map,ArrayList<String> curPath, ArrayList<ArrayList<String>> paths) {if (myNode == null) {paths.add(curPath);return;}curPath.add(0, myNode.val);if (!myNode.pre.isEmpty()) {for (MyNode prevNode : myNode.pre) {getPaths(prevNode, map, new ArrayList<String>(curPath), paths);}} else {getPaths(null, map, curPath, paths);}}boolean reachEnd(String left, String right) {if (left.equals(right)) {return true;}return false;}String getOneDiff(String str, int pos, char c) {StringBuffer sb = new StringBuffer(str);sb.setCharAt(pos, c);return sb.toString();}}class MyNode {String val;int distance;LinkedList<MyNode> pre;MyNode(String v, int distance) {this.val = v;this.distance = distance;pre = new LinkedList<MyNode>();}void addPre(MyNode p) {pre.add(p);}}

不得不说,这个题做的确实痛苦,但是做AC了之后感觉很爽!

也不知道各位大神有什么更好的办法,反正我做的时候确实感觉到很挑战,但很有意思.

过瘾!