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)