N-Queens I & II leetcode

lpy1990

N-Queens I

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.."]
]

回溯法

思路

还是回溯问题, 对于每一个行的每一个格都测试, 如果合理就继续下一行, 不合理就回溯. 这里用一个长度n的数组来存储每一行的情况.
eg:

A[] = [1,0,2,3]表达含义是:
     当前4个皇后所在坐标点为:[[0,1],[1,0],[2,2],[3,3]]
     相当于:A[0] = 1, A[1] = 0, A[2] = 2, A[3] = 3 

这样一来回溯时候就不用对数组进行修改了.

复杂度

时间O(n!) 空间O(n)

代码

public int totalNQueens(int n) {
    if (n <= 0) {
        return 0;
    }
    int[] visit = new int[n];
    int[] res = new int[1];
    helper(res, n, visit, 0);
    return res[0];
}
public void helper(int[] res, int n, int[] visit, int pos) {
    if (pos == n) {
        res[0]++;
        return;
    }
    for(int i = 0; i < n; i++) {
        visit[pos] = i;
        if(isValid(visit, pos)) {
            helper(res, n, visit, pos + 1);
        }
    }
}
public boolean isValid(int[] visit, int pos) {
    for (int i = 0; i < pos; i++) {
        if (visit[i] == visit[pos] || pos - i == Math.abs(visit[pos] - visit[i])) {
            return false;
        }
    }
    return true;
}

N-Queens II

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number
of distinct solutions.

回溯法

思路

与I相同, 只不过不需要返回结果, 而返回结果数量. 所以维护一个数组(不能用int 因为java是按引用传递)

时间O(n!) 空间O(n)

代码

public List<List<String>> solveNQueens(int n) {
    List<List<String>> res = new ArrayList<List<String>>();
    if (n <= 0) {
        return res;
    }
    int[] visit = new int[n];
    helper(res, n, visit, 0);
    return res;
}
public void helper(List<List<String>> res, int n, int[] visit, int pos) {
    if (pos == n) {
        List<String> tem = new ArrayList<String>();
        for (int i = 0; i < n; i++) {
            StringBuilder sb = new StringBuilder();
            for (int j = 0; j < n; j++) {
                if (visit[i] == j) {
                    sb.append("Q");
                } else {
                    sb.append(".");
                }
            }
            tem.add(sb.toString());
        }
        res.add(tem);
        return;
    }
    for(int i = 0; i < n; i++) {
        visit[pos] = i;
        if(isValid(visit, pos)) {
            helper(res, n, visit, pos + 1);
        }
    }
}
public boolean isValid(int[] visit, int pos) {
    for (int i = 0; i < pos; i++) {
        if (visit[i] == visit[pos] || pos - i == Math.abs(visit[pos] - visit[i])) {
            return false;
        }
    }
    return true;
} 
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