​LeetCode刷题实战425：单词方块

示例 1：

输入：
["area","lead","wall","lady","ball"]
输出：
[
[ "wall",
"area",
"lead",
"lady"
],
[ "ball",
"area",
"lead",
"lady"
]
]

解释：
输出包含两个单词方块，输出的顺序不重要，只需要保证每个单词方块内的单词顺序正确即可。
 

示例 2：
输入：
["abat","baba","atan","atal"]

输出：
[
[ "baba",
"abat",
"baba",
"atan"
],
[ "baba",
"abat",
"baba",
"atal"
]
]

解释：
输出包含两个单词方块，输出的顺序不重要，只需要保证每个单词方块内的单词顺序正确即可。

class Solution {
public:
    int limit;
    unordered_map<string,unordered_set<int>> mp;
    vector<vector<string>> wordSquares(vector<string>& words) {
        if(words.empty() or words[0].empty()){return {};}
        vector<vector<string>> res;
        limit=words[0].size();
        vector<int> cur;
        for(int i=0;i            string prefix="";
            for(int j=0;j                prefix+=words[i][j];
                mp[prefix].insert(i);
            }
        }
        for(int i=0;i            cur.emplace_back(i);
            dfs(res,words,cur);
            cur.pop_back();
        }
        return res;
    }

    void dfs(vector<vector<string>>& res,vector<string>& words,vector<int>& cur){
        // for(int x:cur){
        // cout<
        // }cout<<"-----------------"<
        if(cur.size()>=limit){//加入结果数组
            vector<string> cur_res;
            for(int i=0;i                cur_res.emplace_back(words[cur[i]]);
            }
            res.emplace_back(move(cur_res));
            return;
        }
        string prefix="";//先算下一个单词需要满足的前缀
        for(int j=0;j            prefix+=words[cur[j]][cur.size()];
        }
        if(mp.count(prefix)){
            for(int i:mp[prefix]){
                cur.push_back(i);
                dfs(res,words,cur);
                cur.pop_back();
            }
        }
    }
};