37. 解数独

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

  1. 数字 1-9 在每一行只能出现一次。
  2. 数字 1-9 在每一列只能出现一次。
  3. 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 1:

复制代码
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
java 复制代码
    public boolean solveSudoku(char[][] board) {
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                如果为空格就依次把1-9填入
                if (board[i][j] == '.') {
                    for (char k = '1'; k <= '9'; k++) {
                        // 判断填入的数字是否合法
                        if (isValid(i, j, k, board)) {
                            board[i][j] = k;
                            // 递归
                            boolean result = solveSudoku(board);
                            if(result == true)  
                                return true;
                            board[i][j]='.';  //不合法就回溯,回溯前先重新赋值为空
                        }
                    }
                    // 填写完成后返回
                    return false;
                }
            }
        }
        return true;
    }

    /**
     * 判断放入棋盘的数字是否合法
     */
    public boolean isValid(int row, int col, char val, char[][] board) {
        // 验证一行
        for (int i = 0; i < 9; i++) { // 判断行里是否重复
            if (board[row][i] == val) {
                return false;
            }
        }

        // 验证一列
        for (int j = 0; j < 9; j++) { // 判断行里是否重复
            if (board[j][col] == val) {
                return false;
            }
        }

        //    验证3*3
        int startRow = (row / 3) * 3;
        int startCol = (col / 3) * 3;
        for (int i = startRow; i < startRow + 3; i++) { // 判断9方格里是否重复
            for (int j = startCol; j < startCol + 3; j++) {
                if (board[i][j] == val ) {
                    return false;
                }
            }
        }
        return true;
    }
相关推荐
a111776几秒前
高斯泼溅 (Gaussian Splatting) 的 Three.js 实现
开发语言·javascript·ecmascript
代码北人生8 分钟前
agent时代,我们都低估了这个 23k Star 的 Claude Code Skills 项目!
javascript
成都渲染101云渲染66669 分钟前
云渲染全面支持3dsMax 2027,高效渲染体验升级
开发语言·前端·javascript
zs宝来了16 分钟前
微前端架构:qiankun 沙箱隔离与样式冲突
前端·javascript·框架
M ? A32 分钟前
Vue 的 scoped 样式穿透 React 不支持?用 VuReact 编译就行
前端·javascript·vue.js·react.js·面试·开源·vureact
zs宝来了34 分钟前
Vue 3 Composition API:响应式系统与依赖追踪
前端·javascript·框架
向往着的青绿色1 小时前
Java反序列化漏洞(持续更新中)
java·开发语言·计算机网络·安全·web安全·网络安全·网络攻击模型
村上小树1 小时前
非常简单地学习一下slate.js的原理
前端·javascript
web打印社区1 小时前
[特殊字符] 开源好物:web-print-pdf,让 Web 打印像调用接口一样简单!
前端·javascript·vue.js·electron
小短腿的代码世界1 小时前
Qt跨进程通信在交易系统中的应用:让策略引擎与风控模块在毫秒级握手
开发语言·qt