标签:main border overflow [] .com conf tac tom ica
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.."] ]原题链接:https://oj.leetcode.com/problems/n-queens/
题目:n*n的棋盘上放n个皇后,使得不论什么两个皇后不互相攻击。
思路:递归+回溯。
import java.util.ArrayList;
import java.util.List;
public class NQueens {
public static void main(String[] args) {
List<String[]> result = new NQueens().solveNQueens(5);
for(String[] li:result){
for(String str : li){
System.out.println(str);
}
System.out.println("----------------");
}
}
public List<String[]> solveNQueens(int n) {
List<String[]> result = new ArrayList<String[]>();
List<Integer> cols = new ArrayList<Integer>();
if(n <= 0)
return result;
search(result,cols,n);
return result;
}
public void search(List<String[]> result,List<Integer> cols,int n){
if(cols.size() == n){
result.add(draw(cols));
return;
}
for(int col=0;col<n;col++){
if(!isValid(cols,col))
continue;
cols.add(col);
search(result,cols,n);
cols.remove(cols.size()-1);
}
}
public String[] draw(List<Integer> cols){
String[] chess = new String[cols.size()];
for(int i=0;i<chess.length;i++){
chess[i] = "";
for(int j=0;j<cols.size();j++){
if(j==cols.get(i))
chess[i] += "Q";
else
chess[i] += ".";
}
}
return chess;
}
public boolean isValid(List<Integer> cols,int col){
int row = cols.size();
for(int i=0;i<row;i++){
if(cols.get(i) == col)
return false;
if(i - cols.get(i) == row - col)
return false;
if(i+cols.get(i) == row+col)
return false;
}
return true;
}
}
http://huntfor.iteye.com/blog/2039013
标签:main border overflow [] .com conf tac tom ica
原文地址:http://www.cnblogs.com/gavanwanggw/p/7069467.html