【动态规划】【字典树】C++算法：472 连接词

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【动态规划】458:可怜的小猪

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动态规划字典树

LeetCode472 连接词

给你一个不含重复单词的字符串数组 words ，请你找出并返回 words 中的所有连接词。
连接词定义为：一个完全由给定数组中的至少两个较短单词（不一定是不同的两个单词）组成的字符串。
示例 1：
输入：words = [“cat”,“cats”,“catsdogcats”,“dog”,“dogcatsdog”,“hippopotamuses”,“rat”,“ratcatdogcat”]
输出：[“catsdogcats”,“dogcatsdog”,“ratcatdogcat”]
解释：“catsdogcats” 由 “cats”, “dog” 和 “cats” 组成;
“dogcatsdog” 由 “dog”, “cats” 和 “dog” 组成;
“ratcatdogcat” 由 “rat”, “cat”, “dog” 和 “cat” 组成。
示例 2：
输入：words = [“cat”,“dog”,“catdog”]
输出：[“catdog”]
提示：
1 <= words.length <= 10⁴
1 <= words[i].length <= 30
words[i] 仅由小写英文字母组成。
words 中的所有字符串都是唯一的。
1 <= sum(words[i].length) <= 10⁵

动态规划+字典树

分以下几步：
一，将所有单词存到字典树中。
二，枚举各单词s=word[i]，如果s[0,j)是字典树的叶子节点，则将j放到dp[i]中。
三，对dp[i]排序，出掉重复。
四，交换pre,dp。
五，如果pre[i]存在words[i].length，删除之。
六，循环处理pre，直到pre[i]全部为空。
七, for(j ;pre[i]) 如果字典树中存在words[i][j,k)且是叶子节点。将k加到dp[i]中。
八，对dp[i]排序，出掉重复。
九，交换pre,dp。
十,如果pre[i]存在wrods[i].length，将words[i]放到结果中，并清空pre[i]。

封装的字典树

template<class TData=char, int iTypeNum = 26, TData cBegin = 'a'>
class CTrieNode
{
public:CTrieNode* AddChar(TData ele,int& iMaxID){
#ifdef _DEBUGif ((ele < cBegin) || (ele >= cBegin + iTypeNum)){return nullptr;}
#endifconst int index = ele - cBegin;auto ptr = m_vPChilds[ele - cBegin];if (!ptr){m_vPChilds[index] = new CTrieNode();
#ifdef _DEBUGm_vPChilds[index]->m_iID = ++iMaxID;m_childForDebug[ele] = m_vPChilds[index];
#endif}return m_vPChilds[index];}CTrieNode* GetChild(TData ele)const{
#ifdef _DEBUGif ((ele < cBegin) || (ele >= cBegin + iTypeNum)){return nullptr;}
#endifreturn m_vPChilds[ele - cBegin];}
protected:
#ifdef _DEBUGint m_iID = -1;std::unordered_map<TData, CTrieNode*> m_childForDebug;
#endif
public:int m_iLeafIndex = -1;
protected:CTrieNode* m_vPChilds[iTypeNum] = { nullptr };
};template<class TData = char,int iTypeNum = 26, TData cBegin = 'a'>
class CTrie
{
public:	int GetLeadCount(){return m_iLeafCount;}template<class IT>int Add(IT begin, IT end){auto pNode = &m_root;for (; begin != end; ++begin){pNode = pNode->AddChar(*begin,m_iMaxID);}if (-1 == pNode->m_iLeafIndex){pNode->m_iLeafIndex = m_iLeafCount++;}return pNode->m_iLeafIndex;}template<class IT>CTrieNode<TData, iTypeNum, cBegin>* Search(IT begin, IT end){auto ptr = &m_root;for (; begin != end; ++begin){ptr = ptr->GetChild(begin);if (nullptr == ptr){return nullptr;}}return ptr;}CTrieNode<TData, iTypeNum, cBegin> m_root;
protected:	int m_iMaxID = 0;int m_iLeafCount = 0;	
};

核心代码

class Solution {
public:vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {CTrie preTrie;for (const auto& s : words){preTrie.Add(s.begin(), s.end());	}vector<vector<int>> pre(words.size(),vector<int>(1));auto Do = [&](){vector<vector<int>> dp(words.size());for (int i = 0; i < pre.size(); i++){for (const auto& j : pre[i]){			auto p1 = &preTrie.m_root;for (int k = j; k < words[i].length(); k++){p1 = p1->GetChild(words[i][k]);if (nullptr == p1){break;}if (-1 != p1->m_iLeafIndex){dp[i].emplace_back(k + 1);}}}sort(dp[i].begin(), dp[i].end());dp[i].erase(std::unique(dp[i].begin(), dp[i].end()), dp[i].end());}pre.swap(dp);};Do();for (int i = 0 ; i < pre.size(); i++ ){if (pre[i].size() && (pre[i].back() == words[i].length())){pre[i].pop_back();}}vector<string> vRet;bool bDo = true;while (bDo){Do();bDo = false;for (int i = 0; i < pre.size(); i++){if (pre[i].size()){bDo = true;}if (pre[i].size() && (pre[i].back() == words[i].length())){vRet.emplace_back(words[i]);pre[i].clear();}}}return vRet;}
};

测试用例

template<class T>
void Assert(const T& t1, const T& t2)
{assert(t1 == t2);
}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]);}
}int main()
{vector<string> words;{Solution sln;words = { "cat","dog","catdog" };auto res = sln.findAllConcatenatedWordsInADict(words);Assert(vector<string>{"catdog"}, res);}{Solution sln;words = { "cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat" };auto res = sln.findAllConcatenatedWordsInADict(words);Assert(vector<string>{"catsdogcats", "dogcatsdog", "ratcatdogcat"}, res);}{Solution sln;words = { 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};auto res = sln.findAllConcatenatedWordsInADict(words);Assert(vector<string>{"rv", "tp", "koat", "po", "mdtrgv", "wr", "mat", "kmm", "ggk", "kk", "fg", "wrxw", "bilt", "ww", "nxbk", "wd", "fd", "tovdtkr", "ordxzm", "tg", "nxqmxr"}, res);}}

优化

b[j] 表示words[i][0,j) 是否能由一个或多个单词拼接。
Do(j) 的功能： b[j]为true，words[i][j,x)能和那些单词匹配。
Do(0)后，b[s.length()] = 0;是因为至少要拼接两次。

class Solution {
public:vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {CTrie preTrie;for (const auto& s : words){preTrie.Add(s.begin(), s.end());	}int b[31] = { 0 };vector<string> vRet;for (const auto& s : words){			auto Do = [&](int inx){auto p1 = &preTrie.m_root;for (int k = inx; k < s.length(); k++){p1 = p1->GetChild(s[k]);if (nullptr == p1){break;}if (-1 != p1->m_iLeafIndex){b[k + 1] = true;}					}};memset(b, 0, sizeof(b));Do(0);b[s.length()] = 0;for (int i = 1; i < s.length(); i++){if (b[i]){Do(i);}}if (b[s.length()]){vRet.emplace_back(s);}}		return vRet;}
};

2023年1月

class Solution {
public:
vector findAllConcatenatedWordsInADict(vector& words) {
std::unordered_set setHas;
for (const auto& s : words)
{
setHas.insert(s);
}
vector vRet;
for (const auto& s : words)
{
if (Test(s, setHas))
{
vRet.push_back(s);
}
}
return vRet;
}
bool Test(const string& s, const std::unordered_set& setHas)
{
std::vector preLens;
preLens.push_back(0);
while (preLens.size())
{
std::vector lens;
for (const auto& pos : preLens)
{
for (int len = 1; pos + len - 1 < s.length(); len++)
{
if (setHas.count(s.substr(pos, len)))
{
if ((0 == pos) && (s.length() == len))
{
continue;
}
if (pos + len == s.length())
{
return true;
}
lens.push_back(pos + len);
}
}
}
preLens.swap(lens);
}

	 return false;}

};

扩展阅读

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我想对大家说的话
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。
子墨子言之：事无终始，无务多业。也就是我们常说的专业的人做专业的事。
如果程序是一条龙，那算法就是他的是睛