Leetcode51 N-Queens
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N-Queens
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.."] ]
Solution1
- 典型的回溯问题。
public class Solution
public List<List<String>> solveNQueens(int n)
List<List<String>> result = new ArrayList<List<String>>();
if(n<=0) return result;
help(n,0,new ArrayList<String>(),result);
return result;
public void help(int n,int row,ArrayList<String> temp,List<List<String>> result)
if(row>=n)
result.add(temp);
return;
for(int col=0;col<n;col++)
boolean flag = false;
for(int i=0;i<temp.size();i++)//遍历已有的所有行
int usedCol = Integer.MIN_VALUE;
String s = temp.get(i);
for(int j=0;j<s.length();j++)//找到该行的皇后所在列
if(s.charAt(j)=='Q')
usedCol = j;
break;
if(col==usedCol||Math.abs(row-i)==Math.abs(col-usedCol))
flag = true;
break;
if(flag) continue;
ArrayList<String> temp1 = new ArrayList<String>(temp);
char[] line = new char[n];
for(int i=0;i<n;i++) line[i]='.';
line[col] = 'Q';
String current = new String(line);
temp1.add(current);
help(n,row+1,temp1,result);
Solution2
- 用一个数组巧妙的记录每行的皇后所在行和列,可以有更简洁的代码:
public class Solution
public List<List<String>> solveNQueens(int n)
List<List<String>> res = new ArrayList<List<String>>();
help(n, 0, new int[n], res);
return res;
public void help(int n, int row, int[] used, List<List<String>> res)
if(row>=n)
ArrayList<String> temp = new ArrayList<String>();
for(int col:used)
char[] s = new char[n];
Arrays.fill(s,'.');
s[col] = 'Q';
temp.add(new String(s));
res.add(temp);
return;
for(int j=0,i=0;j<n;j++)
for(i=0;i<row;i++) if(j==used[i]||Math.abs(i-row)==Math.abs(j-used[i])) break;
if(i!=row) continue;
used[row] = j;
help(n,row+1,used,res);
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