other solutions

37. Sudoku Solver/0ms.go

@@ -0,0 +1,36 @@
+func solveSudoku(board [][]byte) {
+	var marks [9][27]bool
+	for row := range board {
+		for col := range board[row] {
+			num := board[row][col]
+			if num == '.' {
+				continue
+			}
+			marks[num-49][row], marks[num-49][9+col], marks[num-49][18+row/3*3+col/3] = true, true, true
+		}
+	}
+	sudoku(board, marks, 0, 0)
+}
+
+func sudoku(board [][]byte, marks [9][27]bool, row, col int) bool {
+	if row == len(board) {
+		return true
+	}
+	if board[row][col] != '.' {
+		return sudoku(board, marks, row+(col+1)/9, (col+1)%9)
+	}
+
+	for num := 1; num <= 9; num++ {
+		if marks[num-1][row] || marks[num-1][9+col] || marks[num-1][18+row/3*3+col/3] {
+			continue
+		}
+		board[row][col] = byte(num + 48)
+		marks[num-1][row], marks[num-1][9+col], marks[num-1][18+row/3*3+col/3] = true, true, true
+		if sudoku(board, marks, row+(col+1)/9, (col+1)%9) {
+			return true
+		}
+		board[row][col] = '.'
+		marks[num-1][row], marks[num-1][9+col], marks[num-1][18+row/3*3+col/3] = false, false, false
+	}
+	return false
+}

37. Sudoku Solver/4ms.go

@@ -0,0 +1,40 @@
+func solveSudoku(board [][]byte) {
+	solve(board, 0)
+}
+
+func solve(board [][]byte, k int) bool {
+	if k == 81 {
+		return true
+	}
+
+	r, c := k/9, k%9
+
+	if board[r][c] != '.' {
+		return solve(board, k+1)
+	}
+
+	bi, bj := r/3*3, c/3*3
+
+	isValid := func(b byte) bool {
+		for n := 0; n < 9; n++ {
+			if board[r][n] == b ||
+				board[n][c] == b ||
+				board[bi+n/3][bj+n%3] == b {
+				return false
+			}
+		}
+		return true
+	}
+
+	for b := byte('1'); b <= '9'; b++ {
+		if isValid(b) {
+			board[r][c] = b
+			if solve(board, k+1) {
+				return true
+			}
+		}
+	}
+
+	board[r][c] = '.'
+	return false
+}

37. Sudoku Solver/README.org

@@ -0,0 +1,6 @@
+#+TITLE: Sudoku Solver
+
+我的解法看起来就很杂乱。
+4ms 的用一个 k，好很多。而且效率也是一样的。使用 dfs 和递归。
+
+0ms 的没看懂（