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#yyds干货盘点# LeetCode程序员面试金典:八皇后

时间:2023-02-10 16:01:11浏览次数:48  
标签:yyds .. ... int 金典 new queens LeetCode row

题目:

设计一种算法,打印 N 皇后在 N × N 棋盘上的各种摆法,其中每个皇后都不同行、不同列,也不在对角线上。这里的“对角线”指的是所有的对角线,不只是平分整个棋盘的那两条对角线。

注意:本题相对原题做了扩展

示例:

输入:4

输出:[[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]]

解释: 4 皇后问题存在如下两个不同的解法。

[

 [".Q..",  // 解法 1

 "...Q",

 "Q...",

 "..Q."],

 ["..Q.",  // 解法 2

 "Q...",

 "...Q",

 ".Q.."]

]

代码实现:

class Solution {
public List<List<String>> solveNQueens(int n) {
List<List<String>> solutions = new ArrayList<List<String>>();
int[] queens = new int[n];
Arrays.fill(queens, -1);
Set<Integer> columns = new HashSet<Integer>();
Set<Integer> diagonals1 = new HashSet<Integer>();
Set<Integer> diagonals2 = new HashSet<Integer>();
backtrack(solutions, queens, n, 0, columns, diagonals1, diagonals2);
return solutions;
}

public void backtrack(List<List<String>> solutions, int[] queens, int n, int row, Set<Integer> columns, Set<Integer> diagonals1, Set<Integer> diagonals2) {
if (row == n) {
List<String> board = generateBoard(queens, n);
solutions.add(board);
} else {
for (int i = 0; i < n; i++) {
if (columns.contains(i)) {
continue;
}
int diagonal1 = row - i;
if (diagonals1.contains(diagonal1)) {
continue;
}
int diagonal2 = row + i;
if (diagonals2.contains(diagonal2)) {
continue;
}
queens[row] = i;
columns.add(i);
diagonals1.add(diagonal1);
diagonals2.add(diagonal2);
backtrack(solutions, queens, n, row + 1, columns, diagonals1, diagonals2);
queens[row] = -1;
columns.remove(i);
diagonals1.remove(diagonal1);
diagonals2.remove(diagonal2);
}
}
}

public List<String> generateBoard(int[] queens, int n) {
List<String> board = new ArrayList<String>();
for (int i = 0; i < n; i++) {
char[] row = new char[n];
Arrays.fill(row, '.');
row[queens[i]] = 'Q';
board.add(new String(row));
}
return board;
}
}

标签:yyds,..,...,int,金典,new,queens,LeetCode,row
From: https://blog.51cto.com/u_13321676/6049570

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