Question

52. N-Queens II

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 _the number of distinct solutions to the n-queens puzzle_.

 

Example 1:

Input: n = 4
Output: 2
Explanation: There are two distinct solutions to the 4-queens puzzle as shown.

Example 2:

Input: n = 1
Output: 1

 

Constraints:

• 1 <= n <= 9

Solutions

Solution 1: Backtracking

We design a function $dfs(i)$, which represents starting the search from the $i$th row, and the results of the search are added to the answer.

In the $i$th row, we enumerate each column of the $i$th row. If the current column does not conflict with the queens placed before, then we can place a queen, and then continue to search the next row, that is, call $dfs(i + 1)$.

If a conflict occurs, then we skip the current column and continue to enumerate the next column.

To determine whether a conflict occurs, we need to use three arrays to record whether a queen has been placed in each column, each positive diagonal, and each negative diagonal, respectively.

Specifically, we use the $cols$ array to record whether a queen has been placed in each column, the $dg$ array to record whether a queen has been placed in each positive diagonal, and the $udg$ array to record whether a queen has been placed in each negative diagonal.

The time complexity is $O(n!)$, and the space complexity is $O(n)$. Here, $n$ is the number of queens.

class Solution:
  def totalNQueens(self, n: int) -> int:
    def dfs(i: int):
      if i == n:
        nonlocal ans
        ans += 1
        return
      for j in range(n):
        a, b = i + j, i - j + n
        if cols[j] or dg[a] or udg[b]:
          continue
        cols[j] = dg[a] = udg[b] = True
        dfs(i + 1)
        cols[j] = dg[a] = udg[b] = False

    cols = [False] * 10
    dg = [False] * 20
    udg = [False] * 20
    ans = 0
    dfs(0)
    return ans

class Solution {
  private int n;
  private int ans;
  private boolean[] cols = new boolean[10];
  private boolean[] dg = new boolean[20];
  private boolean[] udg = new boolean[20];

  public int totalNQueens(int n) {
    this.n = n;
    dfs(0);
    return ans;
  }

  private void dfs(int i) {
    if (i == n) {
      ++ans;
      return;
    }
    for (int j = 0; j < n; ++j) {
      int a = i + j, b = i - j + n;
      if (cols[j] || dg[a] || udg[b]) {
        continue;
      }
      cols[j] = true;
      dg[a] = true;
      udg[b] = true;
      dfs(i + 1);
      cols[j] = false;
      dg[a] = false;
      udg[b] = false;
    }
  }
}

class Solution {
public:
  int totalNQueens(int n) {
    bitset<10> cols;
    bitset<20> dg;
    bitset<20> udg;
    int ans = 0;
    function<void(int)> dfs = [&](int i) {
      if (i == n) {
        ++ans;
        return;
      }
      for (int j = 0; j < n; ++j) {
        int a = i + j, b = i - j + n;
        if (cols[j] || dg[a] || udg[b]) continue;
        cols[j] = dg[a] = udg[b] = 1;
        dfs(i + 1);
        cols[j] = dg[a] = udg[b] = 0;
      }
    };
    dfs(0);
    return ans;
  }
};

func totalNQueens(n int) (ans int) {
  cols := [10]bool{}
  dg := [20]bool{}
  udg := [20]bool{}
  var dfs func(int)
  dfs = func(i int) {
    if i == n {
      ans++
      return
    }
    for j := 0; j < n; j++ {
      a, b := i+j, i-j+n
      if cols[j] || dg[a] || udg[b] {
        continue
      }
      cols[j], dg[a], udg[b] = true, true, true
      dfs(i + 1)
      cols[j], dg[a], udg[b] = false, false, false
    }
  }
  dfs(0)
  return
}

function totalNQueens(n: number): number {
  const cols: boolean[] = Array(10).fill(false);
  const dg: boolean[] = Array(20).fill(false);
  const udg: boolean[] = Array(20).fill(false);
  let ans = 0;
  const dfs = (i: number) => {
    if (i === n) {
      ++ans;
      return;
    }
    for (let j = 0; j < n; ++j) {
      let [a, b] = [i + j, i - j + n];
      if (cols[j] || dg[a] || udg[b]) {
        continue;
      }
      cols[j] = dg[a] = udg[b] = true;
      dfs(i + 1);
      cols[j] = dg[a] = udg[b] = false;
    }
  };
  dfs(0);
  return ans;
}