[LeetCode]N-Queens 八皇后问题扩展（经典深层搜索）

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.."]
]

參考：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
;
diag = new boolean[2*n];
anti_diag = new boolean[2*n];
C = new int
;//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
;
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;
}
}
}