leetcode BFS

1. word ladder

 class Solution
 {
 public:
     int ladderLength(string beginWord, string endWord, unordered_set<string> &wordDict)
     {
         queue<string> q;
         q.push(beginWord);
         unordered_map<string, int> umap;
         umap[beginWord] = ;
         for (q.push(beginWord); !q.empty(); q.pop())
         {//这里用for的好处是这本身是一个模块化的过程：先入队；当队非空时进行循环，同时每个循环结束时都要把当前元素出队。用for不容易遗漏q.pop()。
             string word = q.front();
             int step = umap[word] + ;
             for (int i = ; i<word.size(); i++)
             {
                 for (char c = 'a'; c <= 'z'; c++)
                 {
                     if (word[i] != c)
                     {
                         char tmp = word[i];
                         word[i] = c;
                         if (word == endWord)//this line should here, not in the if statement below,coz endWord may not be in dict
                             return step;
                         if (wordDict.find(word) != wordDict.end() && umap.find(word) == umap.end())
                         {
                             umap[word] = step;
                             q.push(word);
                         }
                         word[i] = tmp;//don't forget to restore
                     }
                 }
             }
         }
         return ;
     }
 };