LeetCode 37. Sudoku Solver
LeetCode 37. Sudoku Solver (解数独)
题目
链接
https://leetcode.cn/problems/sudoku-solver/
问题描述
编写一个程序,通过填充空格来解决数独问题。
数独的解法需 遵循如下规则:
数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。(请参考示例图)
数独部分空格内已填入了数字,空白格用 '.' 表示。
示例
输入:board = [["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[".",".",".",".","8",".",".","7","9"]]
输出:[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解释:输入的数独如上图所示,唯一有效的解决方案如下所示:
提示
board.length == 9
board[i].length == 9
board[i][j] 是一位数字或者 '.'
题目数据 保证 输入数独仅有一个解
思路
二重循环,每次放一个数,如果这个位置放9个数都不行,则无解,如果放完了,那就得出唯一解。
复杂度分析
时间复杂度 O(n2)
空间复杂度 O(n)
代码
Java
public static void solveSudoku(char[][] board) {
trace(board);
}
public static boolean trace(char[][] board) {
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
if (board[i][j] != '.') {
continue;
}
for (char k = '1'; k <= '9'; k++) {
if (test(board, i, j, k)) {
board[i][j] = k;
if (trace(board)) {
return true;
}
board[i][j] = '.';
}
}
return false;
}
}
return true;
}
public static boolean test(char[][] board, int x, int y, char value) {
for (int i = 0; i < 9; i++) {
if (board[i][y] == value) {
return false;
}
}
for (int i = 0; i < 9; i++) {
if (board[x][i] == value) {
return false;
}
}
int row = (x / 3) * 3;
int col = (y / 3) * 3;
for (int i = row; i < row + 3; i++) {
for (int j = col; j < col + 3; j++) {
if (board[i][j] == value) {
return false;
}
}
}
return true;
}