n 皇后问题研究的是如何将 n 个皇后放置在 n×n 的棋盘上,并且使皇后彼此之间不能相互攻击。给定一个整数 n,返回 n 皇后不同的解决方案的数量。提示:皇后,是国际象棋中的棋子,意味着国王的妻子。皇后只做一件事,那就是“吃子”。当她遇见可以吃的棋子时,就迅速冲上去吃掉棋子。当然,她横、竖、斜都可走一或 N-1 步,可进可退。
objectNQueenII{ defmain(args: Array[String]): Unit = { println(totalNQueens(4)) }
deftotalNQueens(n: Int): Int = { val board = newArray[Array[Char]](n) for (i <- 0 until n) { board(i) = newArray[Char](n) for (j <- 0 until n) { board(i)(j) = '.' } }
val res = Counter(0) backtrack(board, 0, res) res.get() }
defbacktrack(board: Array[Array[Char]], i: Int, res: Counter): Unit = { if (i >= board.length) { res.increase() return } for (j <- 0 until board.length) { if (isValid(board, i, j)) { board(i)(j) = 'Q' backtrack(board, i + 1, res) board(i)(j) = '.' } } }
defisValid(array: Array[Array[Char]], i: Int, j: Int): Boolean = { //检查同一列是否冲突 for (l <- 0 until array.length) { if (l != i && array(l)(j) == 'Q') { returnfalse } }
var tmpi = i var tmpj = j //检查左上角是否冲突 while ((i > j && tmpj >= 0) || (i <= j && tmpi >= 0)) { if (tmpi != i && array(tmpi)(tmpj) == 'Q') { returnfalse } tmpi -= 1 tmpj -= 1 }
tmpi = i tmpj = j //检查右上角是否冲突 while (tmpj < array.length && tmpi >= 0) { if (tmpi != i && array(tmpi)(tmpj) == 'Q') { returnfalse } tmpi -= 1 tmpj += 1 }