no match for ‘operator=’ 等号两端 不匹配

const auto new_states = state_extend_function(word,dict,visited,end);
unordered_set<string>::iterator itv;
for ( itv=new_states.begin();itv != new_states.end();itv++ ){
string state=*itv;// 操作state
}

报错：

Solution.h:72:15: error: no match for ‘operator=’ (operand types are ‘std

等号两端 不匹配,  可能是auto的C++11版本不同的原因。改成下面的通过：

unordered_set<string> new_states = state_extend_function(word,dict,visited,end);

。。原题 及 修正后的完整代码如下：。。。。。

给出两个单词（start和end）和一个字典，找出所有从start到end的最短转换序列
比如：

每次只能改变一个字母。

变换过程中的中间单词必须在字典中出现。

注意事项

1.所有单词具有相同的长度。

2.所有单词都只包含小写字母。

样例

给出数据如下：
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]
返回
[
    ["hit","hot","dot","dog","cog"],
    ["hit","hot","lot","log","cog"]
  ]

思路：BFS。

#include <iostream>
#include <vector>
#include <limits.h>
#include <string>
#include <queue>
#include <unordered_set>
#include <unordered_map>
using namespace std;

void gen_path(unordered_map<string, vector<string> > &father,
vector<string> &path, const string &start, const string &word,
vector<vector<string> > &result) {
path.push_back(word);
if (word == start) {
result.push_back(path);
reverse(result.back().begin(), result.back().end());
} else {
vector<string> ::iterator itfv;// for (const auto& f : father[word]) {
for (itfv = (father[word]).begin();itfv !=(father[word]).end();itfv++ ) {
auto f=*itfv;
gen_path(father, path, start, f, result);
}
}
path.pop_back();
}

unordered_set<string> state_extend_function(const string &s,
const unordered_set<string> &dict, unordered_set<string> visited,string end) {
unordered_set<string> result;
for (size_t i = 0; i < s.size(); ++i) {
string new_word(s);
for (char c = 'a'; c <= 'z'; c++) {
if (c == new_word[i]) continue;
swap(c, new_word[i]);
if ((dict.count(new_word) > 0 || string(new_word) == string(end) ) &&
!visited.count(new_word)) {
result.insert(new_word);
}
swap(c, new_word[i]); // 恢复该单词
}
}
return result;
}

vector<vector<string> > findLadders(string start, string end,
const unordered_set<string> &dict) {
unordered_set<string> current, next; // 当前层，下一层，用集合是为了去重
unordered_set<string> visited; // 判重
unordered_map<string, vector<string> > father; // 树
bool found = false;
auto state_is_target = [&](const string &s) {
return s == end;
};

current.insert(start);
while (!current.empty() && !found) {
// 先将本层全部置为已访问，防止同层之间互相指向for(const auto& word:current)
unordered_set<string>::iterator it;
for ( it=current.begin();it != current.end();it++  ){
string word=*it;
visited.insert(word);
}

unordered_set<string>::iterator itc;
for ( itc=current.begin();itc != current.end();itc++  ){
string word=*itc;
unordered_set<string> new_states = state_extend_function(word,dict,visited,end);
unordered_set<string>::iterator itv;
for ( itv=new_states.begin();itv != new_states.end();itv++  ){
string state=*itv;

if (state_is_target(state)) found = true;
next.insert(state);
father[state].push_back(word);
// visited.insert(state); // 移动到最上面了
}
}
current.clear();
swap(current, next);
}
vector<vector<string> > result;
if (found) {
vector<string> path;
gen_path(father, path, start, end, result);
}
return result;
}

void main(){
string start="hit",end="cog";
string sArr[]={"hot","dot","dog","lot","log"};
int len=sizeof(sArr)/sizeof(sArr[0] );
unordered_set<string> dict( sArr ,sArr+len);
findLadders(start, end,dict);
cout<<"out="<<endl;
system("pause");
}

。。。

很多高级语言里引入了lambda表达式的概念，即匿名函数。

。。。