标签：style blog http io ar color sp for on
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.."] ]
这里需要用到回溯（su）算法
回溯算法
1.定义一个解空间，即存放一个解的空间
2.开始遍历，DFS，如果没有一个合适的，回退上一步
这个还要多了解，多谢谢才okay
1 public class Solution { 2 int solution[]; 3 int num_queen; 4 List<String[]> result = new ArrayList<String[]>(); 5 6 public List<String[]> solveNQueens(int n) { 7 solution = new int[n + 1]; 8 num_queen = n; 9 placeQueen(1); 10 return result; 11 } 12 13 /**放第n个皇后 14 * @param n 15 */ 16 public void placeQueen(int n){ 17 if(n > num_queen){ 18 String temp[] = new String[num_queen]; 19 String tempStr = ""; 20 21 for(int i = 1; i <= num_queen; i++){ 22 for(int j = 1; j <= num_queen; j++){ 23 if(j == solution[i]) 24 tempStr += "Q"; 25 else 26 tempStr += "."; 27 } 28 temp[i - 1] = tempStr; 29 tempStr = ""; 30 } 31 result.add(temp); 32 }else{ 33 for(int i = 1; i <= num_queen; i++){ 34 solution[n] = i; //将n个皇后放到i列 35 if(isValid(n)) //如果i列有效 36 placeQueen(n + 1); //放下一个皇后 37 } 38 } 39 } 40 41 /** 42 * 第k个皇后的位置是否有效 43 * @param k 44 * @return 45 */ 46 public boolean isValid(int k){ 47 for(int i = 1; i < k; i++){ 48 if(solution[i] == solution[k] || Math.abs(i - k) == Math.abs(solution[i] - solution[k])) 49 return false; 50 } 51 return true; 52 } 53 54 }
标签：style blog http io ar color sp for on
原文地址：http://www.cnblogs.com/luckygxf/p/4119590.html
