LeetCode 题解（95）: Word Ladder II

题目：

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) fromstart toend, 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

题解：

先用BFS生成由start至end的hash table。

再用DFS递归查找所有路径。

第一步的核心数据结构是HashMap<String, HashSet<String>>，表示由String可以通过改变一个字符所能转化的所有字符Set。

注意替换字符再在set中查找的优化，否则超时。

C++版：

class Solution {
public:
    vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) {
       vector<vector<string>> results; 
		if(start.length() == 0 || end.length() == 0 || dict.size() == 0)
			return results;
		dict.insert(end);
		dict.insert(start);
		unordered_map<string, unordered_set<string>> trace;
		for(auto i : dict) {
		    unordered_set<string> temp;
		    trace.insert(pair<string, unordered_set<string>>(i, temp));
		}
		unordered_set<string> q1, q2, visited;
		q1.insert(end);
		bool found = false;
		while(q1.size() != 0 && !found) {
			for(auto i : q1)
			    visited.insert(i);
			for(auto current : q1) {
				for(int i = 0; i < current.length(); i++) {
					for(char j = 'a'; j <= 'z'; j++) {
						string temp = current;
						temp[i] = j;
						if(visited.find(temp) == visited.end() && dict.find(temp) != dict.end()) {
						    if(temp == start)
						        found = true;
						    q2.insert(temp);
						    trace[temp].insert(current);
						}
					}
				}
			}
			q1 = q2;
			q2.clear();
		}
		
		vector<string> result;
		if(found)
			findPaths(trace, result, results, start);
		
		return results;
    }
	
	void findPaths(unordered_map<string, unordered_set<string>>& trace, vector<string>& result, vector<vector<string>>& results, string& start) {
		vector<string> extendedResult = result;
		extendedResult.push_back(start);
		if(trace[start].size() == 0) {
			results.push_back(extendedResult);
			return;
		}
		for(auto i : trace[start]) {
			findPaths(trace, extendedResult, results, i);
		}
	}
};

Java版：

public class Solution {
    
    public List<List<String>> findLadders(String start, String end, Set<String> dict) {
        List<List<String>> results = new ArrayList<List<String>>();
        if(start.isEmpty() || end.isEmpty() || dict.isEmpty())
            return results;
        Set<String> q1 = new HashSet<>();
        Map<String, Set<String>> p = new HashMap<>();
        q1.add(end);
        dict.add(end);
        dict.add(start);
        for(String i : dict) {
            Set<String> temp = new HashSet<>();
            p.put(i, temp);
        }
        Set<String> visited = new HashSet<>();
        boolean found = false;
        while(!q1.isEmpty() && !found) {
            for(String i : q1)
                visited.add(i);
            Set<String> q2 = new HashSet<>();
            for(String current : q1) {
                char[] curChar = current.toCharArray();
                for(int i = 0; i < current.length(); i++) {
                    char original = curChar[i];
                    for(char j = 'a'; j <= 'z'; j++) {
                        curChar[i] = j;
                        String newStr = new String(curChar);
                        if(!visited.contains(newStr) && dict.contains(newStr)) {
                            if(newStr.equals(start))
                                found = true;
                            p.get(newStr).add(current);
                            q2.add(newStr);
                        }
                    }
                    curChar[i] = original;
                }
            }
            q1 = q2;
        }
        
        List<String> result = new ArrayList<>();
        if(found)
            generateResult(result, start, p, results);
        
        return results;
    }
    
    void generateResult(List<String> result, String start, Map<String, Set<String>> p, List<List<String>> results) {
        List<String> extendedResult = new ArrayList<>(result);
        extendedResult.add(start);
        if(p.get(start).size() == 0) {
            results.add(extendedResult);
            return;
        }
        for(String s : p.get(start)) 
            generateResult(extendedResult, s, p, results);
    }
}

Python版：

class Solution:
    # @param start, a string
    # @param end, a string
    # @param dict, a set of string
    # @return a list of lists of string
    def findLadders(self, start, end, dict):
        dict.add(start)
        dict.add(end)
        results, result, q1, visited, found, d = [], [], [end], set([end]), False, {word : [] for word in dict}
        if len(start) == 0 or len(end) == 0 or len(dict) == 0:
            return results

        while len(q1) != 0 and not found:
            for i in q1:
                visited.add(i)
            q2 = set([])
            for current in q1:
                for i in range(len(current)):
                    for j in "abcdefghijklmnopqrstuvwxyz":
                        candidate = current[0:i] + j + current[i+1:]
                        if candidate not in visited and candidate in dict:
                            if candidate == start:
                                found = True
                            q2.add(candidate)
                            d[candidate].append(current)
            q1 = q2

        if found:
            self.findPaths(results, result, d, start)
        return results

    def findPaths(self, results, result, d, start):
        extendedResult = copy.copy(result)
        extendedResult.append(start)
        if not d[start]:
            results.append(extendedResult)
            return
        for i in d[start]:
            self.findPaths(results, extendedResult, d, i)