题目：链接
Given two words (start and end), and a dictionary, find the length of shortest transformation sequence 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"]
As one shortest transformation is"hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length5.
Note:
-
- Return 0 if there is no such transformation sequence.
- All words have the same length.
- All words contain only lowercase alphabetic characters.
解题思路：
显然是搜索问题。
首先，对于start根节点，我们可以逐位枚举26个字母，在字典中搜索其可能的下一步路径。对于已经搜索过的单词，我们期望后续不再处理，这样可以简化搜索空间。因为本题是求最短步长，所以利用广度优先的话，对于已经遍历过的单词，我们可以直接从字典中删除，这样后续即使有可能transformate到它，步长也肯定大于等于之前遍历到它的叔叔or兄弟节点，肯定不是最短步长，可以不考虑该种情况。而用深度优先的话，不能随便删除，举个例子，起点abc的儿子abe和abd，abe先入栈，经过搜索路径abe-aye-tye-tyd-xyd-xyz到达终点xyz，abe的子孙搜索结束之后，沿途经过的节点被删除；搜索C的子孙时，路径abd-xbd-xyd-xyz将不会被搜索到，得出的结果不正确。
同时用curr_min_steps变量记录当前已算出的最小步长，perhaps_min_steps变量为不考虑字典内容时理论上的最小值，对结果进行一定的剪枝。
class Solution {
public:
int ladderLength(string start, string end, unordered_set &dict) {
//一些特殊情况
if(start == end)
return 1;
if(dict.empty())
return 0;
int perhaps_min_steps = 1, curr_min_steps = 10000;
for(int i = 0; i < start.size(); i++){
if(start[i] != end[i]) perhaps_min_steps += 1;
}
if(perhaps_min_steps == 2)
return 2;
queue > path;
path.push(make_pair(start, 1));
while(!path.empty())
{
pair top = path.front();
path.pop();
if(top.second >= curr_min_steps) continue;
string pre = "", post = top.first;
for(int i = 0; i < top.first.size(); i++){
pre = top.first.substr(0, i);
post = post.substr(1);
for(char c = 'a'; c <= 'z'; c++){
string target = pre + c + post;
if(target == end){ //找到了一种
if(top.second + 1 < curr_min_steps) //小于当前最短路径
curr_min_steps = top.second + 1;
if(curr_min_steps == perhaps_min_steps){
return curr_min_steps; //理论上的最小值
} else {
break;
}
}
else{ //中间结果
if(dict.find(target) != dict.end()){
if(top.second+1 > curr_min_steps){
continue;
} else {
path.push(make_pair(target, top.second+1));
dict.erase(target);
}
} else {
continue;
}
}
}
}
}
if(curr_min_steps == 10000)
return 0;
return curr_min_steps;
}
};
进阶版：求所有可能的最短路径
题目：链接
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.
解题思路：
延续上题思路，原来的pair中存储的是词与其祖先的步长，现在修改为祖先的vector即可，不可在push之后就从字典删除节点，因为可能有多条路径共享节点，如red-ted-tex-tax，red-rex-tex-tax。这样修改之后测试，系统提示内存溢出。
确实这样存储vector是比较浪费空间的，我们还需要进行剪枝与删除。
a. 删除：起初先从以下角度进行删除：遍历到某节点时，它的父亲节点可从字典中删除。这样之后报了段错误，提示堆栈溢出。虽然这个提示毫无卵用，但是还是尝试优化了一下：在遍历到第N层节点时，第N-1层的节点均可从字典中删除。
b. 剪枝：记录一些无路可走的节点
其实还有一些优化思路没有实现，比如兄弟节点的祖先路径是相同的，可以考虑邻接存储，减少存储空间。
断断续续做了几个小时，最终提交的准确率只有47%，但是看了一下错误用例，结果是相同的，只是vector的顺序不同，懒得搞了，源码附上。
vector > findLadders(string start, string end, unordered_set &dict) {
vector > result;
if(start == end || dict.empty())
return result;
int curr_min_steps = 10000;
unordered_set badwords;
queue > > path;
vector prepath;
path.push(make_pair(start, prepath));
int level = 0;
vector uncles;
while(!path.empty())
{
pair > top = path.front();
path.pop();
if(top.second.size() == level){ //新的一层第一个节点
level += 1;
for(int i = 0; i < uncles.size(); i++){
dict.erase(uncles[i]);
}
uncles.clear();
}
string pre = "", post = top.first;
int prelength = top.second.size();
if(prelength > curr_min_steps){
break;
}
for(int i = 0; i < top.first.size(); i++){
pre = top.first.substr(0, i);
post = post.substr(1);
for(char c = 'a'; c <= 'z'; c++){
string target = pre + c + post;
if(target == end){ //找到了一种
if(prelength <= curr_min_steps){ //小于当前最短路径
curr_min_steps = prelength;
vector newpath(top.second);
newpath.push_back(top.first);
newpath.push_back(end);
result.push_back(newpath);
break;
} else {
continue;
}
}
else{ //中间结果
if(dict.find(target) != dict.end()
&& find(top.second.begin(), top.second.end(), target) == top.second.end())
{
if(prelength > curr_min_steps){
continue;
} else {
vector newpath(top.second);
newpath.push_back(top.first);
path.push(make_pair(target, newpath));
uncles.push_back(target);
}
} else {
continue;
}
}
}
}
}
return result;
}
提示的错误用例：
测试用例: "qa","sq",["si","go","se","cm","so","ph","mt","db","mb","sb","kr","ln","tm","le","av","sm","ar","ci","ca","br","ti","ba","to","ra","fa","yo","ow","sn","ya","cr","po","fe","ho","ma","re","or","rn","au","ur","rh","sr","tc","lt","lo","as","fr","nb","yb","if","pb","ge","th","pm","rb","sh","co","ga","li","ha","hz","no","bi","di","hi","qa","pi","os","uh","wm","an","me","mo","na","la","st","er","sc","ne","mn","mi","am","ex","pt","io","be","fm","ta","tb","ni","mr","pa","he","lr","sq","ye"] 对应输出应该为: [["qa","ba","be","se","sq"],["qa","ba","bi","si","sq"],["qa","ba","br","sr","sq"],["qa","ca","cm","sm","sq"],["qa","ca","co","so","sq"],["qa","la","ln","sn","sq"],["qa","la","lt","st","sq"],["qa","ma","mb","sb","sq"],["qa","pa","ph","sh","sq"],["qa","ta","tc","sc","sq"],["qa","fa","fe","se","sq"],["qa","ga","ge","se","sq"],["qa","ha","he","se","sq"],["qa","la","le","se","sq"],["qa","ma","me","se","sq"],["qa","na","ne","se","sq"],["qa","ra","re","se","sq"],["qa","ya","ye","se","sq"],["qa","ca","ci","si","sq"],["qa","ha","hi","si","sq"],["qa","la","li","si","sq"],["qa","ma","mi","si","sq"],["qa","na","ni","si","sq"],["qa","pa","pi","si","sq"],["qa","ta","ti","si","sq"],["qa","ca","cr","sr","sq"],["qa","fa","fr","sr","sq"],["qa","la","lr","sr","sq"],["qa","ma","mr","sr","sq"],["qa","fa","fm","sm","sq"],["qa","pa","pm","sm","sq"],["qa","ta","tm","sm","sq"],["qa","ga","go","so","sq"],["qa","ha","ho","so","sq"],["qa","la","lo","so","sq"],["qa","ma","mo","so","sq"],["qa","na","no","so","sq"],["qa","pa","po","so","sq"],["qa","ta","to","so","sq"],["qa","ya","yo","so","sq"],["qa","ma","mn","sn","sq"],["qa","ra","rn","sn","sq"],["qa","ma","mt","st","sq"],["qa","pa","pt","st","sq"],["qa","na","nb","sb","sq"],["qa","pa","pb","sb","sq"],["qa","ra","rb","sb","sq"],["qa","ta","tb","sb","sq"],["qa","ya","yb","sb","sq"],["qa","ra","rh","sh","sq"],["qa","ta","th","sh","sq"]] 你的输出为: [["qa","ba","be","se","sq"],["qa","ba","bi","si","sq"],["qa","ba","br","sr","sq"],["qa","ca","ci","si","sq"],["qa","ca","cm","sm","sq"],["qa","ca","co","so","sq"],["qa","ca","cr","sr","sq"],["qa","fa","fe","se","sq"],["qa","fa","fm","sm","sq"],["qa","fa","fr","sr","sq"],["qa","ga","ge","se","sq"],["qa","ga","go","so","sq"],["qa","ha","he","se","sq"],["qa","ha","hi","si","sq"],["qa","ha","ho","so","sq"],["qa","la","le","se","sq"],["qa","la","li","si","sq"],["qa","la","ln","sn","sq"],["qa","la","lo","so","sq"],["qa","la","lr","sr","sq"],["qa","la","lt","st","sq"],["qa","ma","mb","sb","sq"],["qa","ma","me","se","sq"],["qa","ma","mi","si","sq"],["qa","ma","mn","sn","sq"],["qa","ma","mo","so","sq"],["qa","ma","mr","sr","sq"],["qa","ma","mt","st","sq"],["qa","na","nb","sb","sq"],["qa","na","ne","se","sq"],["qa","na","ni","si","sq"],["qa","na","no","so","sq"],["qa","pa","pb","sb","sq"],["qa","pa","ph","sh","sq"],["qa","pa","pi","si","sq"],["qa","pa","pm","sm","sq"],["qa","pa","po","so","sq"],["qa","pa","pt","st","sq"],["qa","ra","rb","sb","sq"],["qa","ra","re","se","sq"],["qa","ra","rh","sh","sq"],["qa","ra","rn","sn","sq"],["qa","ta","tb","sb","sq"],["qa","ta","tc","sc","sq"],["qa","ta","th","sh","sq"],["qa","ta","ti","si","sq"],["qa","ta","tm","sm","sq"],["qa","ta","to","so","sq"],["qa","ya","yb","sb","sq"],["qa","ya","ye","se","sq"],["qa","ya","yo","so","sq"]]