标签:实现 make func bsp lse ++ else backtrack count
八皇后都成梗了,实际上就是个递归还有对角线公式。
func isNotUnderAttack(row, col, n int, rows, hills, dales []int) bool { res := rows[col] + hills[row - col + 2 * n] + dales[row + col] return res == 0 } func backtrack(row, count, n int, rows, hills, dales []int) int { for col := 0; col < n; col++ { if isNotUnderAttack(row, col, n, rows, hills, dales) { rows[col] = 1 hills[row - col + 2 * n] = 1 dales[row + col] = 1 if row+1 == n { count++ } else { count = backtrack(row + 1, count, n, rows, hills, dales) } rows[col] = 0 hills[row - col + 2 * n] = 0 dales[row + col] = 0 } } return count } func totalNQueens(n int) int { rows := make([]int, n) hills := make([]int, 4*n-1) dales := make([]int, 2*n-1) return backtrack(0, 0, n, rows, hills, dales) }
的确没啥人用Go写
标签:实现 make func bsp lse ++ else backtrack count
原文地址:https://www.cnblogs.com/CherryTab/p/12416500.html