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