leetcode-51 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.

Example:

Input: 4
Output: [
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.

题目要求就是使其两个皇后之间不能互相攻击，即任意两个皇后都不能处于同一行、同一列或同一斜线上，问有多少种摆法

方法一：最简单粗暴的方法：每一行肯定只有一个皇后，并且在每放置一个皇后在不能放置皇后的位置上都用.放置进行遍历

public List<List<String>> solveNQueens(int n) {
   List<List<String>> res = new ArrayList<>();
   if(n<=0) {
       return res;
   }
   String[][] a =new String[n][n];
   getRes(a,res,0,0,n);
   return res;
}


   public void getRes(String[][] a, List<List<String>> res, int begin, int numQueen, int n) {
       if (numQueen == n && n!=0) {
           List<String> tmp = new ArrayList<>();
           for (int i = 0; i < a.length; i++) {
               StringBuffer tmpRes= new StringBuffer();
               for (int j = 0; j < a[0].length; j++) {
                   tmpRes.append(a[i][j]);
               }
               tmp.add(tmpRes.toString());
           }
           res.add(tmp);
           return;
       }
       for (int i = 0; i < a[begin].length; i++) {
           if(begin == 0) {
               a =new String[n][n];
           }
           if ( ".".equals(a[begin][i])) {
               continue;
           }
           String[][] tmp = new String[n][n];
           for (int index = 0; index < a.length; index++) {
               for (int j = 0; j < a[0].length; j++) {
                   tmp[index][j] = a[index][j];
               }
           }
           tmp[begin][i] = "Q";

           //同一横线都除此之外都改为.
           int k=0;
           while(k<tmp[begin].length) {
               if(k != i) {
                   tmp[begin][k] = ".";
               }
               k++;
           }

           //同一竖线都除此之外都改为.
           k=0;
           while(k<tmp.length) {
               if(k != begin) {
                   tmp[k][i] = ".";
               }
               k++;
           }

           int tmpA= begin;
           int tmpB= i;
           while(--tmpA>=0 && --tmpB>=0) {
               tmp[tmpA][tmpB]=".";
           }

           tmpA= begin;
           tmpB= i;
           while(++tmpA<n && ++tmpB<n) {
               tmp[tmpA][tmpB]=".";
           }

           tmpA= begin;
           tmpB= i;
           while(++tmpA<n && --tmpB>=0) {
               tmp[tmpA][tmpB]=".";
           }

           tmpA= begin;
           tmpB= i;
           while(--tmpA>=0 && ++tmpB<n) {
               tmp[tmpA][tmpB]=".";
           }




           getRes(tmp,res,begin+1,numQueen+1,n);
       }

}

方法二：依次放置每行的皇后。在放置的时候，要保持当前的状态为合法，即当前放置位置的同一行、同一列、两条对角线上都不存在皇后。

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