[leetcode]52. N-Queens II N皇后

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 the number of distinct solutions to the n-queens puzzle.

Example:

Input: 4
Output: 2
Explanation: There are two distinct solutions to the 4-queens puzzle as shown below.
[
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]

题意：

code

 class Solution {
    private int count; // 解的个数
     // 这三个变量用于剪枝
     private boolean[] columns;   // 表示已经放置的皇后占据了哪些列
     private boolean[] main_diag; // 占据了哪些主对角线
     private boolean[] anti_diag; // 占据了哪些副对角线

     public int totalNQueens(int n) {
         this.count = 0;
         this.columns = new boolean[n];
         this.main_diag = new boolean[2 * n - 1];
         this.anti_diag = new boolean[2 * n - 1];

         int[] C = new int[n]; // C[i]表示第i行皇后所在的列编号
         dfs(C, 0);
         return this.count;
     }

     void dfs(int[] C, int row) {
         final int N = C.length;
         if (row == N) { // 终止条件，也是收敛条件，意味着找到了一个可行解
             ++this.count;
             return;
         }

         for (int j = 0; j < N; ++j) {  // 扩展状态，一列一列的试
             final boolean ok = !columns[j] &&
                     !main_diag[row - j + N - 1] &&
                     !anti_diag[row + j];
             if (!ok) continue;  // 剪枝：如果合法，继续递归
             // 执行扩展动作
             C[row] = j;
             columns[j] = main_diag[row - j + N - 1] =
                     anti_diag[row + j] = true;
             dfs(C, row + 1);
             // 撤销动作
             // C[row] = -1;
             columns[j] = main_diag[row - j + N - 1] =
                     anti_diag[row + j] = false;
         }
     }
 }