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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.."] ]
solution:经典八皇后问题（这里是n），用递归算法解决
ps:
国际象棋中，皇后会攻击横行竖行和斜线的位置的棋子。楼主用的二维数组比较直观。
package leetcode2; import java.util.*; public class N_QUEEN { public static ArrayList<String[]> Nqueen(int n){ ArrayList<String[]> res=new ArrayList<String[]>(); char[][] location=new char[n][n]; for(int i=0;i<n;i++){ for(int j=0;j<n;j++){ location[i][j]=‘.‘; } } int count=0; Queens(res,n,0,count,location); System.out.print("\t"+res.size()+" \n"); return res; } public static void Queens(ArrayList<String[]> res,int n,int deep,int count,char[][] location1){ if(n==0){ String[] sub=new String[location1.length]; for(int i=0;i<location1.length;i++){ sub[i]=new String(location1[i]); System.out.print(" "+sub[i]); } res.add(sub); count++; // System.out.print(count); return; } for(int y=0;y<location1.length;y++) { if(isvaild(deep,y,location1)){ location1[y][deep]=‘Q‘; Queens(res,n-1,deep+1,count,location1); location1[y][deep]=‘.‘; } } } public static boolean isvaild(int y, int x,char[][] location) { // TODO Auto-generated method stub boolean flag=true; for(int i = 0; i < location.length; i++) { if(location[i][y] == ‘Q‘ || location[x][i] == ‘Q‘) return false; } //left diagonal for(int i = x, j = y; i >=0 && j>=0; j--, i--) if(location[i][j] == ‘Q‘) return false; for(int i = x, j = y; i <location.length && j>=0; j--, i++) if(location[i][j] == ‘Q‘) return false; return true; } public static void main(String[] args) { // TODO Auto-generated method stub System.out.println(Nqueen(4)); } }
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原文地址：http://www.cnblogs.com/joannacode/p/4395730.html
