C++刷LeetCode（648. 单词替换）【字典树】
本帖最后由 糖逗 于 2021-1-26 10:33 编辑题目描述：
在英语中，我们有一个叫做 词根(root)的概念，它可以跟着其他一些词组成另一个较长的单词——我们称这个词为 继承词(successor)。例如，词根an，跟随着单词 other(其他)，可以形成新的单词 another(另一个)。
现在，给定一个由许多词根组成的词典和一个句子。你需要将句子中的所有继承词用词根替换掉。如果继承词有许多可以形成它的词根，则用最短的词根替换它。
你需要输出替换之后的句子。
示例 1：
输入：dictionary = ["cat","bat","rat"], sentence = "the cattle was rattled by the battery"
输出："the cat was rat by the bat"
示例 2：
输入：dictionary = ["a","b","c"], sentence = "aadsfasf absbs bbab cadsfafs"
输出："a a b c"
示例 3：
输入：dictionary = ["a", "aa", "aaa", "aaaa"], sentence = "a aa a aaaa aaa aaa aaa aaaaaa bbb baba ababa"
输出："a a a a a a a a bbb baba a"
示例 4：
输入：dictionary = ["catt","cat","bat","rat"], sentence = "the cattle was rattled by the battery"
输出："the cat was rat by the bat"
示例 5：
输入：dictionary = ["ac","ab"], sentence = "it is abnormal that this solution is accepted"
输出："it is ab that this solution is ac"
提示：
1 <= dictionary.length <= 1000
1 <= dictionary.length <= 100
dictionary 仅由小写字母组成。
1 <= sentence.length <= 10^6
sentence 仅由小写字母和空格组成。
sentence 中单词的总量在范围 内。
sentence 中每个单词的长度在范围 内。
sentence 中单词之间由一个空格隔开。
sentence 没有前导或尾随空格。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/replace-words
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
class Solution {
public:
//字典树的结构
struct TrieTree{
bool flag;
map<char, TrieTree*> next;
TrieTree(): flag(false){};
};
string replaceWords(vector<string>& dictionary, string sentence) {
//构建前缀树
//先建立根节点
TrieTree* root = new TrieTree();
for(auto word : dictionary){
TrieTree* node = root;//从根节点开始搜索
for(auto cha : word){
if((node -> next).count(cha) == 0){
node -> next = new TrieTree();
}
node = node -> next;
}
node -> flag = true;
}
//搜索前缀返回结果
string res;
int start = 0, end = 0;
int len = sentence.size();
for(int i = 0; i < len; i++){
start = i;
TrieTree* node = root;//从根节点开始找
while(i < len && sentence != ' '){
if(node -> flag == true || (node -> next).count(sentence) == 0)break;
node = node -> next];
i++;
}
if(node -> flag == true){
//找到了
end = i;
//定位到下一个start的位置
while(i < len && sentence != ' ')i++;
}else if((node -> next).count(sentence) == 0){
//没找到
//定位到下一个start的位置
while(i < len && sentence != ' ')i++;
end = i;
}
res += " " + sentence.substr(start, end - start);
}
res.erase(res.begin());
return res;
}
}; {:10_312:}
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