【图论回溯广度优先搜索】126. 单词接龙 II-编程知识

本文涉及知识点

图论回溯深度优先搜索广度优先搜索
图论知识汇总

LeetCode 126. 单词接龙 II

按字典 wordList 完成从单词 beginWord 到单词 endWord 转化，一个表示此过程的转换序列是形式上像 beginWord -> s1 -> s2 -> … -> sk 这样的单词序列，并满足：
每对相邻的单词之间仅有单个字母不同。
转换过程中的每个单词 si（1 <= i <= k）必须是字典 wordList 中的单词。注意，beginWord 不必是字典 wordList 中的单词。
sk == endWord
给你两个单词 beginWord 和 endWord ，以及一个字典 wordList 。请你找出并返回所有从 beginWord 到 endWord 的最短转换序列，如果不存在这样的转换序列，返回一个空列表。每个序列都应该以单词列表 [beginWord, s1, s2, …, sk] 的形式返回。

示例 1：
输入：beginWord = “hit”, endWord = “cog”, wordList = [“hot”,“dot”,“dog”,“lot”,“log”,“cog”]
输出：[[“hit”,“hot”,“dot”,“dog”,“cog”],[“hit”,“hot”,“lot”,“log”,“cog”]]
解释：存在 2 种最短的转换序列：
“hit” -> “hot” -> “dot” -> “dog” -> “cog”
“hit” -> “hot” -> “lot” -> “log” -> “cog”
示例 2：

输入：beginWord = “hit”, endWord = “cog”, wordList = [“hot”,“dot”,“dog”,“lot”,“log”]
输出：[]
解释：endWord “cog” 不在字典 wordList 中，所以不存在符合要求的转换序列。

提示：

1 <= beginWord.length <= 5
endWord.length == beginWord.length
1 <= wordList.length <= 500
wordList[i].length == beginWord.length
beginWord、endWord 和 wordList[i] 由小写英文字母组成
beginWord != endWord
wordList 中的所有单词互不相同

图论

beginWord和wordList对应一个节点，注意beginWord如果和wordList[i]相同，则对应节点也相同。
用哈希映射给单词编号，用字典树也可以。
vDis[i]记录节点i到beginWord的最短路径。
vPre[i]记录i到beginWord的最短路径的倒数第二个节点，如果有多条路径，记录所有路径的倒数第二个节点。
n = wordList.length m= beginWord.length $\sum$ = 26 26个小写字母
时间复杂度：以下三步之和：
一，建立临接表。O(nnm) $\approx$ 10⁶。
二，广度优先，等于边数，边数最多n $\times$ n 。故时间复杂度O(nn), $\approx$ 10⁶。
三，回溯。计算复杂。怀疑是 $\sum^4$ ，即每个节点和endWord相同字符+1，其实不是。如：“hit”,“hot”,“dot”,“dog”,“cog”
hit有三个字符和cog不同，hot dot 有两个字符和cog不同，dog有一个字符和cog不同。

代码

核心代码

class CStrToIndex
{
public:CStrToIndex(const vector<string>& wordList) {for (const auto& str : wordList){Add(str);}}void Add(const string& str){if (m_mIndexs.count(str)) { return; }m_mIndexs[str] = m_strs.size();m_strs.push_back(str);}vector<string> m_strs;int GetIndex(const string& str){if (m_mIndexs.count(str)) { return m_mIndexs[str]; }return -1;}
protected:unordered_map<string, int> m_mIndexs;
};
class Solution {
public:vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {CStrToIndex inx(wordList);inx.Add(beginWord);m_c = inx.m_strs.size();vector<vector<int>> vNeiBo(m_c);for (int i = 0; i < m_c; i++) {for (int j = i + 1; j < m_c; j++) {int iNotSame = 0;for (int k = 0; k < inx.m_strs[i].length(); k++) {iNotSame += (inx.m_strs[i][k] != inx.m_strs[j][k]);}if (1 == iNotSame) {vNeiBo[i].emplace_back(j);vNeiBo[j].emplace_back(i);}}}m_iBegin = inx.GetIndex(beginWord);		m_iEnd = inx.GetIndex(endWord);if (-1 == m_iEnd) { return {}; };queue<int> que;vector<int> dis(m_c,m_c);vector<vector<int>> vPre(m_c);auto Add = [&](int cur, int next) {const int iNew = dis[cur] + 1;if (iNew > dis[next]) { return; }			if (iNew < dis[next]) {vPre[next].clear();dis[next] = iNew;que.emplace(next);}vPre[next].emplace_back(cur);};dis[m_iBegin] = 0;que.emplace(m_iBegin);while (que.size()) {auto cur = que.front();que.pop();for (const auto& next : vNeiBo[cur]) {Add(cur, next);}}BackTrack(m_iEnd, inx, vPre);if (dis[m_iEnd] >= m_c) { return {}; }return m_vRet;}void BackTrack(int cur, CStrToIndex& inx, const vector<vector<int>>& vPre){if (m_iBegin == cur) {m_vCur.emplace_back(cur);m_vRet.emplace_back();for (auto it = m_vCur.rbegin(); it != m_vCur.rend(); ++it) {m_vRet.back().emplace_back(inx.m_strs[*it]);}m_vCur.pop_back();}m_vCur.emplace_back(cur);for (const auto& pre : vPre[cur]){BackTrack(pre,inx, vPre);}m_vCur.pop_back();}vector<vector<string>> m_vRet;vector<int> m_vCur;int m_c, m_iBegin,m_iEnd;
};

测试用例

template<class T>
void Assert(const vector<T>& v1, const vector<T>& v2)
{if (v1.size() != v2.size()){assert(false);return;}for (int i = 0; i < v1.size(); i++){assert(v1[i] == v2[i]);}
}template<class T>
void Assert(const T& t1, const T& t2)
{assert(t1 == t2);
}int main()
{string beginWord, endWord;vector<string> wordList;{Solution slu;beginWord = "red", endWord = "tax", wordList = { "ted","tex","red","tax","tad","den","rex","pee" };auto res = slu.findLadders(beginWord, endWord, wordList);Assert({ {"red","ted","tex","tax"},{"red","rex","tex","tax"},{"red","ted","tad","tax"} }, res);}{Solution slu;beginWord = "hit", endWord = "cog", wordList = { "hot","dot","dog","lot","log" };auto res = slu.findLadders(beginWord, endWord, wordList);Assert({  }, res);}{Solution slu;beginWord = "hit", endWord = "cog", wordList = { "hot","dot","dog","lot","log","cog" };auto res = slu.findLadders(beginWord, endWord, wordList);Assert({ {"hit","hot","dot","dog","cog"},{"hit","hot","lot","log","cog"} }, res);}{Solution slu;beginWord = "a", endWord = "c", wordList = { "a","b","c" };auto res = slu.findLadders(beginWord, endWord, wordList);Assert({ {"a","c"} }, res);}}

2023年4月版

class Solution {
public:vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {if (wordList.end() == std::find(wordList.begin(), wordList.end(), beginWord)){wordList.emplace_back(beginWord);}for (const auto& word : wordList){AddNeib(word);}vector<vector<std::string>> vRet;std::queue<int> preQue;m_vDis.resize(m_vNeib.size());const int iBeginIndex = m_mMaskIndex[StrToMask(beginWord)];m_vDis[iBeginIndex] = 1;preQue.emplace(iBeginIndex);const long long llMask = StrToMask(endWord);if (0 == m_mMaskIndex.count(llMask)){return vRet;}const int iEndIndex = m_mMaskIndex[llMask];for (int i = 1; preQue.size(); i++){std::queue<int> curQue;while (preQue.size()){const auto curIndex = preQue.front();preQue.pop();if (curIndex == iEndIndex){vector<string> strs((i+1)/2);dfs(vRet, strs, iEndIndex, i);return vRet;}for (const auto & next : m_vNeib[curIndex]){if (m_vDis[next]){continue;}m_vDis[next] = i + 1;curQue.emplace(next);}}preQue.swap(curQue);}return vRet;}void dfs(std::vector<std::vector<string>>& vRet, std::vector<string>& strs, int iCurNode, int iCurLeve){if (iCurLeve & 1){strs[(iCurLeve - 1) / 2] = m_vStrs[iCurNode];if (1 == iCurLeve){vRet.emplace_back(strs);return;}}for (const auto& next : m_vNeib[iCurNode]){if (1 + m_vDis[next] != iCurLeve){continue;}dfs(vRet, strs, next, iCurLeve - 1);}}long long StrToMask(const string& s){long long llRet = 0;for (const auto& ch : s){llRet = llRet * m_iUnit + ch - 'a' + 1;}return llRet;}string MaskToStr(long long llMask){vector<char> chas;while (llMask){chas.emplace_back(llMask%m_iUnit - 1 + 'a');llMask /= m_iUnit;}std::reverse(chas.begin(), chas.end());chas.emplace_back(0);return std::string(chas.begin(), chas.end());}int AddWord(const string& s){return AddWord(StrToMask(s));}int AddWord(long long llMask){if (m_mMaskIndex.count(llMask)){return m_mMaskIndex[llMask];}m_vNeib.emplace_back();m_vStrs.emplace_back(MaskToStr(llMask));return m_mMaskIndex[llMask] = m_vNeib.size()-1;}void AddNeib(const string& s){		const long long llMask = StrToMask(s);int index = AddWord(llMask);long long llMul = 1;for (int i = 0; i < s.length(); i++){const char& ch = s[s.length() - 1 - i];auto tmp = llMask - llMul*(ch - 'a' + 1);int index2 = AddWord(tmp);m_vNeib[index].emplace_back(index2);m_vNeib[index2].emplace_back(index);llMul *= m_iUnit;}}std::unordered_map<long long, int> m_mMaskIndex;std::vector<vector<int>> m_vNeib;std::vector<std::string> m_vStrs;vector<int> m_vDis;const int m_iUnit = 27;
};

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我想对大家说的话
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闻缺陷则喜是一个美好的愿望，早发现问题，早修改问题，给老板节约钱。
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