[LeetCode]N-Queens

N-Queens

The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

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.."]
]

回溯算法，参考这个，比较好理解一点。
假设数组为int state[n], state[i]表示第 i 行皇后所在的列。那么在新的一行 k 放置一个皇后后:

• 判断列是否冲突，只需要看state数组中state[0…k-1] 是否有和state[k]相等；
• 判断对角线是否冲突：如果两个皇后在同一对角线，那么|row1-row2| = |column1 - column2|，（row1，column1），（row2，column2）分别为冲突的两个皇后的位置

class Solution {
private:
  vector<vector<string> > res;
public:
  vector<vector<string> > solveNQueens(int n) {
    vector<int> state(n, -1);
    helper(state, 0);
    return res;
  }
  void helper(vector<int> &state, int row)
  {//放置第row行的皇后
    int n = state.size();
    if(row == n)
    {
      vector<string>tmpres(n, string(n,'.'));
      for(int i = 0; i < n; i++)
        tmpres[i][state[i]] = 'Q';
      res.push_back(tmpres);
      return;
    }
    for(int col = 0; col < n; col++)
      if(isValid(state, row, col))
      {
        state[row] = col;
        helper(state, row+1);
        state[row] = -1;
      }
  }
  
  bool isValid(vector<int> &state, int row, int col)
  {
    for(int i = 0; i < row; i++)
      if(state[i] == col || abs(row - i) == abs(col - state[i]))
        return false;
    return true;
  }
};