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

Given an integer n, return all distinct solutions to the n-queens puzzle. You may return the answer in any order.

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 1:

Input: n = 4
Output: [[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above

Example 2:

Input: n = 1
Output: [["Q"]]

 

 

public static List<List<String>> solveNQueens(int n) {
List<List<String>> res = new ArrayList<>();
int[] queenList = new int[n]; //第i个位置存放的数表示row行时,Q的列,如果queenList[0] = 2,表示第0行的第二列是Q
placeQueen(queenList, 0, n, res);//在第0行放Q
return res;
}

private static void placeQueen(int[] queenList, int row, int n, List<List<String>> res) {
//如果已经填满,就生成结果
if (row == n) {
ArrayList<String> list = new ArrayList<>();
for (int i = 0; i < n; i++) {
String str = "";
for (int col = 0; col < n; col++){
if(queenList[i] == col) {
str += "Q";
} else {
str += ".";
}
}
list.add(str);
}
res.add(list);
}
for (int col = 0; col < n; col++) {//循环每一列
if (isValid(queenList, row, col)) { //如果在该列放入Q不冲突的话
queenList[row] = col;
placeQueen(queenList, row + 1, n, res);
}
}
}

private static boolean isValid(int[] queenList, int row, int col) {
for (int i = 0; i < row; i++) {
int pos = queenList[i];
if (pos == col) { //和新加入的Q处于同一列
return false;
}
if (pos + row - i == col) { //在新加入的Q的右对角线上
return false;
}
if (pos - row + i == col) { //在新加入的Q的左对角线上
return false;
}
}
return true;
}

原文地址:http://www.cnblogs.com/MarkLeeBYR/p/16884995.html

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