bfs/word-ladder-ii
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2023-12-01
Word Ladder II
描述
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
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.
分析
跟 Word Ladder 比，这题是求路径本身，不是路径长度，也是 BFS，略微麻烦点。
求一条路径和求所有路径有很大的不同，求一条路径，每个状态节点只需要记录一个前驱即可；求所有路径时，有的状态节点可能有多个父节点，即要记录多个前驱。
如果当前路径长度已经超过当前最短路径长度，可以中止对该路径的处理，因为我们要找的是最短路径。
单队列
// Word Ladder II
// 时间复杂度O(n)，空间复杂度O(n)
public class Solution {
public List<List<String>> findLadders(String beginWord, String endWord,
Set<String> wordList) {
Queue<String> q = new LinkedList<>();
HashMap<String, Integer> visited = new HashMap<>(); // 判重
HashMap<String, ArrayList<String>> father = new HashMap<>(); // DAG
final Function<String, Boolean> stateIsValid = (String s) ->
wordList.contains(s) || s.equals(endWord);
final Function<String, Boolean> stateIsTarget = (String s) ->
s.equals(endWord);
final Function<String, HashSet<String> > stateExtend = (String s) -> {
HashSet<String> result = new HashSet<>();
char[] array = s.toCharArray();
for (int i = 0; i < array.length; ++i) {
final char old = array[i];
for (char c = 'a'; c <= 'z'; c++) {
// 防止同字母替换
if (c == array[i]) continue;
array[i] = c;
String newState = new String(array);
final int newDepth = visited.get(s) + 1;
if (stateIsValid.apply(newState)) {
if (visited.containsKey(newState)) {
final int depth = visited.get(newState);
if (depth < newDepth) {
// do nothing
} else if (depth == newDepth) {
result.add(newState);
} else {
throw new IllegalStateException("not possible to get here");
}
} else {
result.add(newState);
}
}
array[i] = old; // 恢复该单词
}
}
return result;
};
List<List<String>> result = new ArrayList<>();
q.offer(beginWord);
visited.put(beginWord, 0);
while (!q.isEmpty()) {
String state = q.poll();
// 如果当前路径长度已经超过当前最短路径长度，
// 可以中止对该路径的处理，因为我们要找的是最短路径
if (!result.isEmpty() && (visited.get(state) + 1) > result.get(0).size()) break;
if (stateIsTarget.apply(state)) {
ArrayList<String> path = new ArrayList<>();
genPath(father, beginWord, state, path, result);
continue;
}
// 必须挪到下面，比如同一层A和B两个节点均指向了目标节点，
// 那么目标节点就会在q中出现两次，输出路径就会翻倍
// visited.insert(state);
// 扩展节点
HashSet<String> newStates = stateExtend.apply(state);
for (String newState : newStates) {
if (!visited.containsKey(newState)) {
q.offer(newState);
visited.put(newState, visited.get(state)+1);
}
ArrayList<String> parents = father.getOrDefault(newState, new ArrayList<>());
parents.add(state);
father.put(newState, parents);
}
}
return result;
}
private static void genPath(HashMap<String, ArrayList<String>> father,
String start, String state, List<String> path,
List<List<String>> result) {
path.add(state);
if (state.equals(start)) {
if (!result.isEmpty()) {
if (path.size() < result.get(0).size()) {
result.clear();
} else if (path.size() == result.get(0).size()) {
// do nothing
} else {
throw new IllegalStateException("not possible to get here");
}
}
ArrayList<String> tmp = new ArrayList<>(path);
Collections.reverse(tmp);
result.add(tmp);
} else {
for (String f : father.get(state)) {
genPath(father, start, f, path, result);
}
}
path.remove(path.size() - 1);
}
}
// Word Ladder II
// 时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
vector<vector<string> > findLadders(const string& start,
const string& end, const unordered_set<string> &dict) {
queue<string> q;
unordered_map<string, int> visited; // 判重
unordered_map<string, vector<string> > father; // DAG
auto state_is_valid = [&](const string& s) {
return dict.find(s) != dict.end() || s == end;
};
auto state_is_target = [&](const string &s) {return s == end; };
auto state_extend = [&](const string &s) {
unordered_set<string> result;
const int new_depth = visited[s] + 1;
for (size_t i = 0; i < s.size(); ++i) {
string new_state = s;
for (char c = 'a'; c <= 'z'; c++) {
// 防止同字母替换
if (c == new_state[i]) continue;
swap(c, new_state[i]);
if (state_is_valid(new_state)) {
auto visited_iter = visited.find(new_state);
if (visited_iter != visited.end()) {
const int depth = visited_iter->second;
if (depth < new_depth) {
// do nothing
}
else if (depth == new_depth) {
result.insert(new_state);
}
else { // not possible
throw std::logic_error("not possible to get here");
}
}
else {
result.insert(new_state);
}
}
swap(c, new_state[i]); // 恢复该单词
}
}
return result;
};
vector<vector<string>> result;
q.push(start);
visited[start] = 0;
while (!q.empty()) {
// 千万不能用 const auto&，pop() 会删除元素，
// 引用就变成了悬空引用
const auto state = q.front();
q.pop();
// 如果当前路径长度已经超过当前最短路径长度，
// 可以中止对该路径的处理，因为我们要找的是最短路径
if (!result.empty() && visited[state] + 1 > result[0].size()) break;
if (state_is_target(state)) {
vector<string> path;
gen_path(father, start, state, path, result);
continue;
}
// 必须挪到下面，比如同一层A和B两个节点均指向了目标节点，
// 那么目标节点就会在q中出现两次，输出路径就会翻倍
// visited.insert(state);
// 扩展节点
const auto& new_states = state_extend(state);
for (const auto& new_state : new_states) {
if (visited.find(new_state) == visited.end()) {
q.push(new_state);
visited[new_state] = visited[state] + 1;
}
father[new_state].push_back(state);
}
}
return result;
}
private:
void gen_path(unordered_map<string, vector<string> > &father,
const string &start, const string &state, vector<string> &path,
vector<vector<string> > &result) {
path.push_back(state);
if (state == start) {
if (!result.empty()) {
if (path.size() < result[0].size()) {
result.clear();
}
else if (path.size() == result[0].size()) {
// do nothing
}
else { // not possible
throw std::logic_error("not possible to get here ");
}
}
result.push_back(path);
reverse(result.back().begin(), result.back().end());
}
else {
for (const auto& f : father[state]) {
gen_path(father, start, f, path, result);
}
}
path.pop_back();
}
};
双队列
// Word Ladder II
// 时间复杂度O(n)，空间复杂度O(n)
public class Solution {
public List<List<String>> findLadders(String beginWord, String endWord,
Set<String> wordList) {
// 当前层，下一层，用unordered_set是为了去重，例如两个父节点指向
// 同一个子节点，如果用vector, 子节点就会在next里出现两次，其实此
// 时 father 已经记录了两个父节点，next里重复出现两次是没必要的
HashSet<String> current = new HashSet<>();
HashSet<String> next = new HashSet<>();
HashSet<String> visited = new HashSet<>(); // 判重
HashMap<String, ArrayList<String>> father = new HashMap<>(); // DAG
int level = -1; // 层次
final Function<String, Boolean> stateIsValid = (String s) ->
wordList.contains(s) || s.equals(endWord);
final Function<String, Boolean> stateIsTarget = (String s) ->
s.equals(endWord);
final Function<String, HashSet<String> > stateExtend = (String s) -> {
HashSet<String> result = new HashSet<>();
char[] array = s.toCharArray();
for (int i = 0; i < array.length; ++i) {
final char old = array[i];
for (char c = 'a'; c <= 'z'; c++) {
// 防止同字母替换
if (c == array[i]) continue;
array[i] = c;
String newState = new String(array);
if (stateIsValid.apply(newState) &&
!visited.contains(newState)) {
result.add(newState);
}
array[i] = old; // 恢复该单词
}
}
return result;
};
List<List<String>> result = new ArrayList<>();
current.add(beginWord);
while (!current.isEmpty()) {
++ level;
// 如果当前路径长度已经超过当前最短路径长度，
// 可以中止对该路径的处理，因为我们要找的是最短路径
if (!result.isEmpty() && level + 1 > result.get(0).size()) break;
// 1. 延迟加入visited, 这样才能允许两个父节点指向同一个子节点
// 2. 一股脑current 全部加入visited, 是防止本层前一个节点扩展
// 节点时，指向了本层后面尚未处理的节点，这条路径必然不是最短的
for (String state : current)
visited.add(state);
for (String state : current) {
if (stateIsTarget.apply(state)) {
ArrayList<String> path = new ArrayList<>();
genPath(father, beginWord, state, path, result);
continue;
}
// 扩展节点
HashSet<String> newStates = stateExtend.apply(state);
for (String newState : newStates) {
next.add(newState);
ArrayList<String> parents = father.getOrDefault(newState, new ArrayList<>());
parents.add(state);
father.put(newState, parents);
}
}
current.clear();
// swap
HashSet<String> tmp = current;
current = next;
next = tmp;
}
return result;
}
private static void genPath(HashMap<String, ArrayList<String>> father,
String start, String state, List<String> path,
List<List<String>> result) {
path.add(state);
if (state.equals(start)) {
if (!result.isEmpty()) {
if (path.size() < result.get(0).size()) {
result.clear();
} else if (path.size() == result.get(0).size()) {
// do nothing
} else {
throw new IllegalStateException("not possible to get here");
}
}
ArrayList<String> tmp = new ArrayList<>(path);
Collections.reverse(tmp);
result.add(tmp);
} else {
for (String f : father.get(state)) {
genPath(father, start, f, path, result);
}
}
path.remove(path.size() - 1);
}
}
// Word Ladder II
// 时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
vector<vector<string> > findLadders(const string& start,
const string& end, const unordered_set<string> &dict) {
// 当前层，下一层，用unordered_set是为了去重，例如两个父节点指向
// 同一个子节点，如果用vector, 子节点就会在next里出现两次，其实此
// 时 father 已经记录了两个父节点，next里重复出现两次是没必要的
unordered_set<string> current, next;
unordered_set<string> visited; // 判重
unordered_map<string, vector<string> > father; // DAG
int level = -1; // 层次
auto state_is_valid = [&](const string& s) {
return dict.find(s) != dict.end() || s == end;
};
auto state_is_target = [&](const string &s) {return s == end;};
auto state_extend = [&](const string &s) {
unordered_set<string> result;
for (size_t i = 0; i < s.size(); ++i) {
string new_word(s);
for (char c = 'a'; c <= 'z'; c++) {
// 防止同字母替换
if (c == new_word[i]) continue;
swap(c, new_word[i]);
if (state_is_valid(new_word) &&
visited.find(new_word) == visited.end()) {
result.insert(new_word);
}
swap(c, new_word[i]); // 恢复该单词
}
}
return result;
};
vector<vector<string> > result;
current.insert(start);
while (!current.empty()) {
++ level;
// 如果当前路径长度已经超过当前最短路径长度，可以中止对该路径的
// 处理，因为我们要找的是最短路径
if (!result.empty() && level+1 > result[0].size()) break;
// 1. 延迟加入visited, 这样才能允许两个父节点指向同一个子节点
// 2. 一股脑current 全部加入visited, 是防止本层前一个节点扩展
// 节点时，指向了本层后面尚未处理的节点，这条路径必然不是最短的
for (const auto& state : current)
visited.insert(state);
for (const auto& state : current) {
if (state_is_target(state)) {
vector<string> path;
gen_path(father, path, start, state, result);
continue;
}
const auto new_states = state_extend(state);
for (const auto& new_state : new_states) {
next.insert(new_state);
father[new_state].push_back(state);
}
}
current.clear();
swap(current, next);
}
return result;
}
private:
void gen_path(unordered_map<string, vector<string> > &father,
vector<string> &path, const string &start, const string &word,
vector<vector<string> > &result) {
path.push_back(word);
if (word == start) {
if (!result.empty()) {
if (path.size() < result[0].size()) {
result.clear();
result.push_back(path);
} else if(path.size() == result[0].size()) {
result.push_back(path);
} else {
// not possible
throw std::logic_error("not possible to get here");
}
} else {
result.push_back(path);
}
reverse(result.back().begin(), result.back().end());
} else {
for (const auto& f : father[word]) {
gen_path(father, path, start, f, result);
}
}
path.pop_back();
}
};
图的广搜
前面的解法，在状态扩展的时候，每次都是从'a'到'z'全部枚举一遍，重复计算，比较浪费，其实当给定字典dict后，单词与单词之间的路径就固定下来了，本质上单词与单词之间构成了一个无向图。如果事先把这个图构建出来，那么状态扩展就会大大加快。
import java.util.*;
import java.util.function.Predicate;
import java.util.function.Function;
// Word Ladder II
// 时间复杂度O(n)，空间复杂度O(n)
public class Solution {
public List<List<String>> findLadders(String beginWord, String endWord,
Set<String> wordList) {
Queue<String> q = new LinkedList<>();
HashMap<String, Integer> visited = new HashMap<>(); // 判重
HashMap<String, ArrayList<String>> father = new HashMap<>(); // DAG
// only used by stateExtend()
final HashMap<String, HashSet<String>> g = buildGraph(wordList);
final Function<String, Boolean> stateIsValid = (String s) ->
wordList.contains(s) || s.equals(endWord);
final Function<String, Boolean> stateIsTarget = (String s) ->
s.equals(endWord);
final Function<String, List<String> > stateExtend = (String s) -> {
List<String> result = new ArrayList<>();
final int newDepth = visited.get(s) + 1;
HashSet<String> list = g.get(s);
if (list == null) return result;
for (String newState : list) {
if (stateIsValid.apply(newState)) {
if (visited.containsKey(newState)) {
final int depth = visited.get(newState);
if (depth < newDepth) {
// do nothing
} else if (depth == newDepth) {
result.add(newState);
} else {
throw new IllegalStateException("not possible to get here");
}
} else {
result.add(newState);
}
}
}
return result;
};
List<List<String>> result = new ArrayList<>();
q.offer(beginWord);
visited.put(beginWord, 0);
while (!q.isEmpty()) {
String state = q.poll();
// 如果当前路径长度已经超过当前最短路径长度，
// 可以中止对该路径的处理，因为我们要找的是最短路径
if (!result.isEmpty() && (visited.get(state) + 1) > result.get(0).size()) break;
if (stateIsTarget.apply(state)) {
ArrayList<String> path = new ArrayList<>();
genPath(father, beginWord, state, path, result);
continue;
}
// 必须挪到下面，比如同一层A和B两个节点均指向了目标节点，
// 那么目标节点就会在q中出现两次，输出路径就会翻倍
// visited.insert(state);
// 扩展节点
List<String> newStates = stateExtend.apply(state);
for (String newState : newStates) {
if (!visited.containsKey(newState)) {
q.offer(newState);
visited.put(newState, visited.get(state)+1);
}
ArrayList<String> parents = father.getOrDefault(newState, new ArrayList<>());
parents.add(state);
father.put(newState, parents);
}
}
return result;
}
private static void genPath(HashMap<String, ArrayList<String>> father,
String start, String state, List<String> path,
List<List<String>> result) {
path.add(state);
if (state.equals(start)) {
if (!result.isEmpty()) {
if (path.size() < result.get(0).size()) {
result.clear();
} else if (path.size() == result.get(0).size()) {
// do nothing
} else {
throw new IllegalStateException("not possible to get here");
}
}
ArrayList<String> tmp = new ArrayList<>(path);
Collections.reverse(tmp);
result.add(tmp);
} else {
for (String f : father.get(state)) {
genPath(father, start, f, path, result);
}
}
path.remove(path.size() - 1);
}
private static HashMap<String, HashSet<String>> buildGraph(Set<String> dict) {
HashMap<String, HashSet<String>> adjacency_list = new HashMap<>();
for (String word: dict) {
char[] array = word.toCharArray();
for (int i = 0; i < array.length; ++i) {
final char old = array[i];
for (char c = 'a'; c <= 'z'; c++) {
// 防止同字母替换
if (c == array[i]) continue;
array[i] = c;
String newWord = new String(array);
if (dict.contains(newWord)) {
HashSet<String> list = adjacency_list.getOrDefault(
word, new HashSet<>());
list.add(newWord);
adjacency_list.put(word, list);
}
array[i] = old; // 恢复该单词
}
}
}
return adjacency_list;
}
}
// Word Ladder II
// 时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
vector<vector<string> > findLadders(const string& start,
const string& end, const unordered_set<string> &dict) {
queue<string> q;
unordered_map<string, int> visited; // 判重
unordered_map<string, vector<string> > father; // DAG
// only used by state_extend()
const unordered_map<string, unordered_set<string> >& g = build_graph(dict);
auto state_is_valid = [&](const string& s) {
return dict.find(s) != dict.end() || s == end;
};
auto state_is_target = [&](const string &s) {return s == end; };
auto state_extend = [&](const string &s) {
vector<string> result;
const int new_depth = visited[s] + 1;
auto iter = g.find(s);
if (iter == g.end()) return result;
const auto& list = iter->second;
for (const auto& new_state : list) {
if (state_is_valid(new_state)) {
auto visited_iter = visited.find(new_state);
if (visited_iter != visited.end()) {
const int depth = visited_iter->second;
if (depth < new_depth) {
// do nothing
}
else if (depth == new_depth) {
result.push_back(new_state);
} else { // not possible
throw std::logic_error("not possible to get here");
}
}
else {
result.push_back(new_state);
}
}
}
return result;
};
vector<vector<string>> result;
q.push(start);
visited[start] = 0;
while (!q.empty()) {
// 千万不能用 const auto&，pop() 会删除元素，
// 引用就变成了悬空引用
const auto state = q.front();
q.pop();
// 如果当前路径长度已经超过当前最短路径长度，
// 可以中止对该路径的处理，因为我们要找的是最短路径
if (!result.empty() && visited[state] + 1 > result[0].size()) break;
if (state_is_target(state)) {
vector<string> path;
gen_path(father, start, state, path, result);
continue;
}
// 必须挪到下面，比如同一层A和B两个节点均指向了目标节点，
// 那么目标节点就会在q中出现两次，输出路径就会翻倍
// visited.insert(state);
// 扩展节点
const auto& new_states = state_extend(state);
for (const auto& new_state : new_states) {
if (visited.find(new_state) == visited.end()) {
q.push(new_state);
visited[new_state] = visited[state] + 1;
}
father[new_state].push_back(state);
}
}
return result;
}
private:
void gen_path(unordered_map<string, vector<string> > &father,
const string &start, const string &state, vector<string> &path,
vector<vector<string> > &result) {
path.push_back(state);
if (state == start) {
if (!result.empty()) {
if (path.size() < result[0].size()) {
result.clear();
}
else if (path.size() == result[0].size()) {
// do nothing
}
else { // not possible
throw std::logic_error("not possible to get here ");
}
}
result.push_back(path);
reverse(result.back().begin(), result.back().end());
}
else {
for (const auto& f : father[state]) {
gen_path(father, start, f, path, result);
}
}
path.pop_back();
}
unordered_map<string, unordered_set<string> > build_graph(
const unordered_set<string>& dict) {
unordered_map<string, unordered_set<string> > adjacency_list;
for (const auto& word : dict) {
for (size_t i = 0; i < word.size(); ++i) {
string new_word(word);
for (char c = 'a'; c <= 'z'; c++) {
// 防止同字母替换
if (c == new_word[i]) continue;
swap(c, new_word[i]);
if ((dict.find(new_word) != dict.end())) {
auto iter = adjacency_list.find(word);
if (iter != adjacency_list.end()) {
iter->second.insert(new_word);
}
else {
adjacency_list.insert(pair<string,
unordered_set<string >> (word, unordered_set<string>()));
adjacency_list[word].insert(new_word);
}
}
swap(c, new_word[i]); // 恢复该单词
}
}
}
return adjacency_list;
}
};