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 = 0; i < n; ++i){
            board[i] = i + 1;//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 = 1; i += index; index++; return i;});
            transform(board.begin(), board.end(), afterSub.begin(), 
　　　　　　　　　　　　　　[](int i)->int{static int index = 1; 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 = 0; i < n; i++){
                      string tmp = "";
                      for(int j = 0; j < n; ++j){
                          tmp.append(1, j == board[i]-1 ? '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(0,n);
           return ret;
    }

    bool check(int pos, int line)
    {
        for(int i = 0; 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15         }
        return true;
    }    

    void dfs(int dep, int maxDep)
    {
        if(dep == maxDep){
            vector<string> tmpVec;
            for(int i = 0; i < maxDep; ++i){
                string tmp = "";
                for(int j = 0; j < maxDep; ++j){
                    tmp.append(1, j == a[i] ? 'Q' : '.');
                }
                tmpVec.push_back(tmp);
                tmp.clear();
            }
            ret.push_back(tmpVec);
            tmpVec.clear();
        }
        for(int i = 0; i < maxDep; ++i){
            if(canUse[i] && check(i, dep)){
                canUse[i] = false;
                a[dep] = i;
                dfs(dep + 1, maxDep);
                canUse[i] = true;
            }
        }
    }
private:
    vector<vector<string>> ret;
    int a[100];
    bool canUse[100];
};