leetcode笔记:51. n

# Title Difficulty Topic
51 N-Queens Hard Backtracking

Description

The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

8-queens

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.

Example:

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Input: 4
Output: [
[".Q..", // Solution 1
"...Q",
"Q...",
"..Q."],

["..Q.", // Solution 2
"Q...",
"...Q",
".Q.."]
]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.

Backtracking

LeetCode 51. N-Queens - 花花酱 刷题找工作 EP41视频里讲的很清晰。

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class  {
public List<List<String>> solveNQueens(int n) {
char[][] board = new char[n][n];
for(int i=0; i<n; i++) {
for(int j=0; j<n; j++) {
board[i][j] = '.';
}
}
List<List<String>> res = new ArrayList<>();
dfs(board, 0, res);
return res;
}

public void dfs(char[][] board, int row, List<List<String>> res) {
int n = board.length;
if(row == n) {
List<String> tmp = new ArrayList<>();
for(int i=0; i<n; i++) {
tmp.add(new String(board[i]));
}
res.add(tmp);
return;
}

for(int j=0; j<n; j++) {
if(isValid(board, row, j)) {
board[row][j] = 'Q';
dfs(board, row+1, res);
board[row][j] = '.';
}
}
}

public boolean isValid(char[][] board, int row, int col) {
int n = board.length;
for(int j=0; j<n; j++) {
if(board[j][col] != '.') return false;
if(row+col-j >= 0 && row+col-j < n && board[row+col-j][j] != '.') return false;
if(row-col+j >= 0 && row-col+j < n && board[row-col+j][j] != '.') return false;
}
return true;
}
}