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] = 1;
        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] + 1;
            for (int i = 0; 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 0;
    }
};