Word Ladder II
Word Ladder II
问题：
Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that:
- Only one letter can be changed at a time
- Each intermediate word must exist in the dictionary
思路：
bfs dfs结合
我的代码：
public class Solution { public List<List<String>> findLadders(String start, String end, Set<String> dict) { if(isValid(start, end)) { List<String> tmpList = new ArrayList<String>(); tmpList.add(start); tmpList.add(end); list.add(tmpList); return list; } Set<String> rst = new HashSet<String>(); for(String str : dict) { if(isValid(end,str)) { rst.add(str); } } List<String> tmpList = new ArrayList<String>(); dfs(tmpList, rst, dict, start, end); List<List<String>> result = new ArrayList<List<String>>(); for(List<String> partList: list) { if(partList.size() == shortPath) result.add(partList); } return result; } private List<List<String>> list = new ArrayList<List<String>>(); private int shortPath = Integer.MAX_VALUE; public void dfs(List<String> tmpList, Set<String> rst, Set<String> dict, String start, String end) { tmpList.add(start); if(rst.contains(start)) { tmpList.add(end); if(shortPath >= tmpList.size()) { list.add(new ArrayList(tmpList)); shortPath = tmpList.size(); } tmpList.remove(tmpList.size()-1); tmpList.remove(tmpList.size()-1); return ; } for(char c = 'a'; c <= 'z'; c++) { for(int j = 0; j < start.length(); j++) { if(start.charAt(j) == c) continue; String str = replace(start, j, c); if(dict.contains(str)) { dict.remove(str); dfs(tmpList, rst, dict, str, end); dict.add(str); } } } tmpList.remove(tmpList.size()-1); } private String replace(String s, int index, char c) { char[] chars = s.toCharArray(); chars[index] = c; return new String(chars); } public boolean isValid(String one, String two) { int count = 0; for(int i = 0; i < one.length(); i++) { if(one.charAt(i) != two.charAt(i)) count++; } return count == 1 ? true : false; } }
他人代码：
public class Solution { public List<List<String>> findLadders(String start, String end, Set<String> dict) { List<List<String>> ladders = new ArrayList<List<String>>(); Map<String, List<String>> map = new HashMap<String, List<String>>(); Map<String, Integer> distance = new HashMap<String, Integer>(); dict.add(start); dict.add(end); bfs(map, distance, start, end, dict); List<String> path = new ArrayList<String>(); dfs(ladders, path, end, start, distance, map); return ladders; } void dfs(List<List<String>> ladders, List<String> path, String crt, String start, Map<String, Integer> distance, Map<String, List<String>> map) { path.add(crt); if (crt.equals(start)) { Collections.reverse(path); ladders.add(new ArrayList<String>(path)); Collections.reverse(path); } else { for (String next : map.get(crt)) { if (distance.containsKey(next) && distance.get(crt) == distance.get(next) + 1) { dfs(ladders, path, next, start, distance, map); } } } path.remove(path.size() - 1); } void bfs(Map<String, List<String>> map, Map<String, Integer> distance, String start, String end, Set<String> dict) { Queue<String> q = new LinkedList<String>(); q.offer(start); distance.put(start, 0); for (String s : dict) { map.put(s, new ArrayList<String>()); } while (!q.isEmpty()) { String crt = q.poll(); List<String> nextList = expand(crt, dict); for (String next : nextList) { map.get(next).add(crt); if (!distance.containsKey(next)) { distance.put(next, distance.get(crt) + 1); q.offer(next); } } } } List<String> expand(String crt, Set<String> dict) { List<String> expansion = new ArrayList<String>(); for (int i = 0; i < crt.length(); i++) { for (char ch = 'a'; ch <= 'z'; ch++) { if (ch != crt.charAt(i)) { String expanded = crt.substring(0, i) + ch + crt.substring(i + 1); if (dict.contains(expanded)) { expansion.add(expanded); } } } } return expansion; } }
学习之处：
- 我的代码只能过小的数据集，大的数据集就超时了，主要是太多的的重复计算非最短路径中，需要先进行BFS确定哪些是最短的路径
- 直接复制的别人的代码AC的，有点浮躁啊，看看了，下次再刷leetcode的时候，再自己写一遍，到了hard的难度了，越来越多的题，hold不住了。
- 改掉自己身上不好的习惯，Day by day, 进步一点点