Question

51. N-Queens

中文文档

Description

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

Given an integer n, return _all distinct solutions to the n-queens puzzle_. You may return the answer in any order.

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

 

Example 1:

Input: n = 4
Output: [[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above

Example 2:

Input: n = 1
Output: [["Q"]]

 

Constraints:

• 1 <= n <= 9

Solutions

Solution 1: DFS (Backtracking)

We define three arrays $col$, $dg$, and $udg$ to represent whether there is a queen in the column, the main diagonal, and the anti-diagonal, respectively. If there is a queen at position $(i, j)$, then $col[j]$, $dg[i + j]$, and $udg[n - i + j]$ are all $1$. In addition, we use an array $g$ to record the current state of the chessboard, where all elements in $g$ are initially '.'.

Next, we define a function $dfs(i)$, which represents placing queens starting from the $i$th row.

In $dfs(i)$, if $i = n$, it means that we have completed the placement of all queens. We put the current $g$ into the answer array and end the recursion.

Otherwise, we enumerate each column $j$ of the current row. If there is no queen at position $(i, j)$, that is, $col[j]$, $dg[i + j]$, and $udg[n - i + j]$ are all $0$, then we can place a queen, that is, change $g[i][j]$ to 'Q', and set $col[j]$, $dg[i + j]$, and $udg[n - i + j]$ to $1$. Then we continue to search the next row, that is, call $dfs(i + 1)$. After the recursion ends, we need to change $g[i][j]$ back to '.' and set $col[j]$, $dg[i + j]$, and $udg[n - i + j]$ to $0$.

In the main function, we call $dfs(0)$ to start recursion, and finally return the answer array.

The time complexity is $O(n^2 \times n!)$, and the space complexity is $O(n)$. Here, $n$ is the integer given in the problem.

class Solution:
  def solveNQueens(self, n: int) -> List[List[str]]:
    def dfs(i: int):
      if i == n:
        ans.append(["".join(row) for row in g])
        return
      for j in range(n):
        if col[j] + dg[i + j] + udg[n - i + j] == 0:
          g[i][j] = "Q"
          col[j] = dg[i + j] = udg[n - i + j] = 1
          dfs(i + 1)
          col[j] = dg[i + j] = udg[n - i + j] = 0
          g[i][j] = "."

    ans = []
    g = [["."] * n for _ in range(n)]
    col = [0] * n
    dg = [0] * (n << 1)
    udg = [0] * (n << 1)
    dfs(0)
    return ans

class Solution {
  private List<List<String>> ans = new ArrayList<>();
  private int[] col;
  private int[] dg;
  private int[] udg;
  private String[][] g;
  private int n;

  public List<List<String>> solveNQueens(int n) {
    this.n = n;
    col = new int[n];
    dg = new int[n << 1];
    udg = new int[n << 1];
    g = new String[n][n];
    for (int i = 0; i < n; ++i) {
      Arrays.fill(g[i], ".");
    }
    dfs(0);
    return ans;
  }

  private void dfs(int i) {
    if (i == n) {
      List<String> t = new ArrayList<>();
      for (int j = 0; j < n; ++j) {
        t.add(String.join("", g[j]));
      }
      ans.add(t);
      return;
    }
    for (int j = 0; j < n; ++j) {
      if (col[j] + dg[i + j] + udg[n - i + j] == 0) {
        g[i][j] = "Q";
        col[j] = dg[i + j] = udg[n - i + j] = 1;
        dfs(i + 1);
        col[j] = dg[i + j] = udg[n - i + j] = 0;
        g[i][j] = ".";
      }
    }
  }
}

class Solution {
public:
  vector<vector<string>> solveNQueens(int n) {
    vector<int> col(n);
    vector<int> dg(n << 1);
    vector<int> udg(n << 1);
    vector<vector<string>> ans;
    vector<string> t(n, string(n, '.'));
    function<void(int)> dfs = [&](int i) -> void {
      if (i == n) {
        ans.push_back(t);
        return;
      }
      for (int j = 0; j < n; ++j) {
        if (col[j] + dg[i + j] + udg[n - i + j] == 0) {
          t[i][j] = 'Q';
          col[j] = dg[i + j] = udg[n - i + j] = 1;
          dfs(i + 1);
          col[j] = dg[i + j] = udg[n - i + j] = 0;
          t[i][j] = '.';
        }
      }
    };
    dfs(0);
    return ans;
  }
};

func solveNQueens(n int) (ans [][]string) {
  col := make([]int, n)
  dg := make([]int, n<<1)
  udg := make([]int, n<<1)
  t := make([][]byte, n)
  for i := range t {
    t[i] = make([]byte, n)
    for j := range t[i] {
      t[i][j] = '.'
    }
  }
  var dfs func(int)
  dfs = func(i int) {
    if i == n {
      tmp := make([]string, n)
      for i := range tmp {
        tmp[i] = string(t[i])
      }
      ans = append(ans, tmp)
      return
    }
    for j := 0; j < n; j++ {
      if col[j]+dg[i+j]+udg[n-i+j] == 0 {
        col[j], dg[i+j], udg[n-i+j] = 1, 1, 1
        t[i][j] = 'Q'
        dfs(i + 1)
        t[i][j] = '.'
        col[j], dg[i+j], udg[n-i+j] = 0, 0, 0
      }
    }
  }
  dfs(0)
  return
}

function solveNQueens(n: number): string[][] {
  const col: number[] = new Array(n).fill(0);
  const dg: number[] = new Array(n << 1).fill(0);
  const udg: number[] = new Array(n << 1).fill(0);
  const ans: string[][] = [];
  const t: string[][] = Array(n)
    .fill(0)
    .map(() => Array(n).fill('.'));
  const dfs = (i: number) => {
    if (i === n) {
      ans.push(t.map(x => x.join('')));
      return;
    }
    for (let j = 0; j < n; ++j) {
      if (col[j] + dg[i + j] + udg[n - i + j] === 0) {
        t[i][j] = 'Q';
        col[j] = dg[i + j] = udg[n - i + j] = 1;
        dfs(i + 1);
        col[j] = dg[i + j] = udg[n - i + j] = 0;
        t[i][j] = '.';
      }
    }
  };
  dfs(0);
  return ans;
}

public class Solution {
  private int n;
  private int[] col;
  private int[] dg;
  private int[] udg;
  private IList<IList<string>> ans = new List<IList<string>>();
  private IList<string> t = new List<string>();

  public IList<IList<string>> SolveNQueens(int n) {
    this.n = n;
    col = new int[n];
    dg = new int[n << 1];
    udg = new int[n << 1];
    dfs(0);
    return ans;
  }

  private void dfs(int i) {
    if (i == n) {
      ans.Add(new List<string>(t));
      return;
    }
    for (int j = 0; j < n; ++j) {
      if (col[j] + dg[i + j] + udg[n - i + j] == 0) {
        char[] row = new char[n];
        Array.Fill(row, '.');
        row[j] = 'Q';
        t.Add(new string(row));
        col[j] = dg[i + j] = udg[n - i + j] = 1;
        dfs(i + 1);
        col[j] = dg[i + j] = udg[n - i + j] = 0;
        t.RemoveAt(t.Count - 1);
      }
    }
  }
}