LeetCode第三十七题-实现数独

最新推荐文章于 2021-02-12 05:23:57 发布

原创最新推荐文章于 2021-02-12 05:23:57 发布 · 164 阅读

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本文介绍了一种使用递归和回溯技术解决数独问题的方法。通过逐个填充空格并验证其有效性，确保每行、每列及每个3x3宫格内的数字1-9恰好出现一次。

Sudoku Solver

问题简介：给定一个大小9*9的二维数组,输入部分数字,其它字符用’.‘代替,要求完成剩余数字的计算即字符’.‘处
数独的要求：
1.每个数字1-9必须在每行中恰好出现一次
2.每个数字1-9必须在每列中恰好出现一次
3.数字1-9中的每一个必须在网格的9个3×3子框中的每一个中恰好出现一次
举例：
输入：
[
[“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”]
]
即：
在这里插入图片描述
结果：填充未完成的部分,即红色数字位置

解法一：
利用递归的思路,逐个填充原字符’.'处的数字,当填充每个数字时进行判断,判断填充的数字是否有效,直到递归所有字符

class Solution {
    public void solveSudoku(char[][] board) {
        solve(board, 0, 0);
    }
    boolean solve(char[][] board, int i, int j) {
        if (i == 9) return true;
        if (j >= 9) return solve(board, i + 1, 0);
        if (board[i][j] == '.') {
            for (int k = 1; k <= 9; ++k) {
                board[i][j] = (char)(k + '0');
                if (isValid(board, i , j)) {
                    if (solve(board, i, j + 1)) return true;
                }
                board[i][j] = '.';
            }
        } else {
            return solve(board, i, j + 1);
        }
        return false;
    }
    boolean isValid(char[][] board, int i, int j) {
        for (int col = 0; col < 9; ++col) {
            if (col != j && board[i][j] == board[i][col]) return false;
        }
        for (int row = 0; row < 9; ++row) {
            if (row != i && board[i][j] == board[row][j]) return false;
        }
        for (int row = i / 3 * 3; row < i / 3 * 3 + 3; ++row) {
            for (int col = j / 3 * 3; col < j / 3 * 3 + 3; ++col) {
                if ((row != i || col != j) && board[i][j] == board[row][col]) return false;
            }
        }
        return true;
}
}

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