feat: Add Ruby code - chapter "Backtracking" (#1373)

* [feat] add ruby code - chapter backtracking

* feat: add ruby code block - chapter backtracking

codes/ruby/chapter_backtracking/n_queens.rb

@@ -0,0 +1,61 @@
+=begin
+File: n_queens.rb
+Created Time: 2024-05-21
+Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
+=end
+
+### 回溯算法：n 皇后 ###
+def backtrack(row, n, state, res, cols, diags1, diags2)
+  # 当放置完所有行时，记录解
+  if row == n
+    res << state.map { |row| row.dup }
+    return
+  end
+
+  # 遍历所有列
+  for col in 0...n
+    # 计算该格子对应的主对角线和次对角线
+    diag1 = row - col + n - 1
+    diag2 = row + col
+    # 剪枝：不允许该格子所在列、主对角线、次对角线上存在皇后
+    if !cols[col] && !diags1[diag1] && !diags2[diag2]
+      # 尝试：将皇后放置在该格子
+      state[row][col] = "Q"
+      cols[col] = diags1[diag1] = diags2[diag2] = true
+      # 放置下一行
+      backtrack(row + 1, n, state, res, cols, diags1, diags2)
+      # 回退：将该格子恢复为空位
+      state[row][col] = "#"
+      cols[col] = diags1[diag1] = diags2[diag2] = false
+    end
+  end
+end
+
+### 求解 n 皇后 ###
+def n_queens(n)
+  # 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
+  state = Array.new(n) { Array.new(n, "#") }
+  cols = Array.new(n, false) # 记录列是否有皇后
+  diags1 = Array.new(2 * n - 1, false) # 记录主对角线上是否有皇后
+  diags2 = Array.new(2 * n - 1, false) # 记录次对角线上是否有皇后
+  res = []
+  backtrack(0, n, state, res, cols, diags1, diags2)
+
+  res
+end
+
+### Driver Code ###
+if __FILE__ == $0
+  n = 4
+  res = n_queens(n)
+
+  puts "输入棋盘长宽为 #{n}"
+  puts "皇后放置方案共有 #{res.length} 种"
+
+  for state in res
+    puts "--------------------"
+    for row in state
+      p row
+    end
+  end
+end