LeetCode OJ：N-Queens（N皇后问题）

Given an integer n, return all distinct solutions to the n-queens puzzle.

Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space respectively.

For example,
There exist two distinct solutions to the 4-queens puzzle:

[
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]

经典的N皇后问题，我首先是用brute force，但是这样做的话一直会TLE：

 class Solution {
 public:
     vector<vector<string>> solveNQueens(int n) {
         vector<int> board(n);
         vector<vector<string>> ret;
         vector<string> tmpVec;
         for(int i = ; i < n; ++i){
             board[i] = i + ;//确保了上下左右不会有相邻的元素
         }
         vector<int> afterAdd(n);
         vector<int> afterSub(n);
         for(;;){
             transform(board.begin(), board.end(), afterAdd.begin(), 
　　　　　　　　　　　　　　[](int i)->int{static int index = ; i += index; index++; return i;});
             transform(board.begin(), board.end(), afterSub.begin(), 
　　　　　　　　　　　　　　[](int i)->int{static int index = ; i -= index; index++; return i;});
               set<int>afterSubSet(afterSub.begin(), afterSub.end());
               set<int>afterAddSet(afterAdd.begin(), afterAdd.end());
               if(afterAddSet.size() == n && afterSubSet.size() == n){
                   for(int i = ; i < n; i++){
                       string tmp = "";
                       for(int j = ; j < n; ++j){
                           tmp.append(, j == board[i]- ? 'Q' : '.');
                       }
                       tmpVec.push_back(tmp);
                       tmp.clear();
                   }
                   ret.push_back(tmpVec);
                   tmpVec.clear();
               }
               if(!next_permutation(board.begin(), board.end())){
                   return ret;
                   break;
               }
         }
     }
 };

后来就只能使用dfs了，代码如下：

 class Solution {
 public:
     vector<vector<string>> solveNQueens(int n) {
            ret.clear();
            memset(canUse, true, sizeof(canUse));
            dfs(,n);
            return ret;
     }

     bool check(int pos, int line)
     {
         for(int i = ; i < line; ++i){
             if(line - i == abs(pos - a[i]))    //这一步很重要
                 return false; //相差n行的两个皇后的位置不可以也正好相差n个，那样一定是不可以的         }
         return true;
     }    

     void dfs(int dep, int maxDep)
     {
         if(dep == maxDep){
             vector<string> tmpVec;
             for(int i = ; i < maxDep; ++i){
                 string tmp = "";
                 for(int j = ; j < maxDep; ++j){
                     tmp.append(, j == a[i] ? 'Q' : '.');
                 }
                 tmpVec.push_back(tmp);
                 tmp.clear();
             }
             ret.push_back(tmpVec);
             tmpVec.clear();
         }
         for(int i = ; i < maxDep; ++i){
             if(canUse[i] && check(i, dep)){
                 canUse[i] = false;
                 a[dep] = i;
                 dfs(dep + , maxDep);
                 canUse[i] = true;
             }
         }
     }
 private:
     vector<vector<string>> ret;
     int a[];
     bool canUse[];
 };