Word Ladder 2 BFS

最新推荐文章于 2024-09-10 03:45:52 发布

JohnWill_

最新推荐文章于 2024-09-10 03:45:52 发布

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分类专栏：广度优先搜索BFS 队列

本文链接：https://blog.youkuaiyun.com/weixin_40039738/article/details/79673497

广度优先搜索BFS 同时被 2 个专栏收录

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本文探讨了一个经典的图搜索问题——寻找两个单词间最短转换路径的方法。通过构建一个由单词构成的图，使用广度优先搜索策略来寻找从起始单词到目标单词的所有最短路径。文章详细介绍了算法实现，并针对一种特定的实现方式进行了讨论，指出其存在的问题（如超时），为读者提供了解决此类问题的基础思路。

题目描述

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.

以下代码超时：

class Solution {
public:
vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) {
vector<vector<string>> result;
vector<string> path;
bool found = false;
unordered_set<string> visited;
unordered_map<string, unordered_set<string>> father;

queue<string> current, next;

if(start.size() != end.size()) return result;
current.push(start);
while(!current.empty())
{
string str=current.front();
current.pop();
string a = str;
for(int i=0; i<str.size(); ++i)
{
char tmp=str[i];
for(char c='a' ;c<='z'; ++c)
{
if(c == tmp) continue;
str[i]=c;
if(str == end)
{
found = true;
break;
}
if(dict.find(str)!= dict.end() && !visited.count(str))
{
visited.insert(str);
next.push(str);
father[str].insert(a);
}
str[i]=tmp;
}
}// end for
if(current.empty() && !found) current=next;
}// end while
buildPath(father, path, start, end, result);
return result;
}
void buildPath( unordered_map<string, unordered_set<string>> &father, vector<string> &path, const string &start, const string &end, vector<vector<string> > &result)
{
path.push_back(end);
if(end == start)
{
result.push_back(path);
reverse(result.back().begin(), result.back().end());
}
else
{
for(auto f : father[end])
buildPath(father, path, start, f, result);
}
path.pop_back();
}
};