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[LeetCode]N-Queens 八皇后问题扩展（经典深搜）
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.."] ]参考：LeetCode 题解
戴方勤 (soulmachine@gmail.com)
https://github.com/soulmachine/leetcode
public class Solution {
boolean[] column = null;
boolean[] diag = null;
boolean[] anti_diag = null;
int[]C = null;
String model = null;
public List<String[]> solveNQueens(int n) {
column = new boolean[n];
diag = new boolean[2*n];
anti_diag = new boolean[2*n];
C = new int[n];//Q在第i行的第几列
List<String []> res = new ArrayList<>();
dfs(0,res,n);
return res;
}
private void dfs(int row,List<String[]> res,int n){
if(row==n){
String [] str = new String[n];
for(int i=0;i<n;i++){
StringBuilder sb = new StringBuilder();
for(int j=0;j<n;j++){
if(C[i]==j){
sb.append("Q");
}else{
sb.append(".");
}
}
str[i] = sb.toString();
}
res.add(str);
return;
}
for(int j=0;j<n;j++){
if(!column[j]&&!anti_diag[row+j]&&!diag[n-row-1+j]){
C[row]=j;
}else{
continue;
}
column[j]=anti_diag[row+j]=diag[n-row-1+j]=true;
dfs(row+1,res,n);
column[j]=anti_diag[row+j]=diag[n-row-1+j]=false;
}
}
}[LeetCode]N-Queens 八皇后问题扩展（经典深搜）
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