Leetcode 36：有效的数独

题目描述

判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。

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。

上图是一个部分填充的有效的数独。

数独部分空格内已填入了数字，空白格用 ‘.’ 表示。

示例 1:

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”]
]
输出: true

示例 2:

输入:
[
[“8”,“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: false
解释: 除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。
但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

说明:

• 一个有效的数独（部分已被填充）不一定是可解的。
• 只需要根据以上规则，验证已经填入的数字是否有效即可。
• 给定数独序列只包含数字 1-9 和字符 ‘.’ 。
• 给定数独永远是 9x9 形式的。

我的解法：利用set特性，将判断分为三个步骤

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        set<char> s;
        //行重复检测
        for (int m = 0; m < 9; m++){
            for (int n = 0; n < 9; n++){
                if (board[m][n] != '.'){
                    if (s.count(board[m][n])){ return false; }
                    else{ s.insert(board[m][n]); }
                }
            }
            s.clear();
        }

        //列重复检测
        for (int n = 0; n < 9; n ++){
            for (int m = 0; m < 9; m++){
                if (board[m][n] != '.'){
                    if (s.count(board[m][n])){ return false; }
                    else{ s.insert(board[m][n]); }
                }
            }
            s.clear();
        }

        int m0, m1, n0, n1;
        //块重复检测
        for (int m = 0; m < 9; m++){
            for (int n = 0; n < 9; n++){
                if (board[m][n] != '.'){
                    if (m % 3 == 0){ m0 = m + 1; m1 = m + 2; }
                    else if (m % 3 == 1){ m0 = m - 1; m1 = m + 1; }
                    else{ m0 = m - 2; m1 = m - 1; }

                    if (n % 3 == 0){ n0 = n + 1; n1 = n + 2; }
                    else if (n % 3 == 1){ n0 = n - 1; n1 = n + 1; }
                    else{ n0 = n - 2; n1 = n - 1; }

                    if (m0 > m){
                        if (board[m0][n0] != '.'){ s.insert(board[m0][n0]); }
                        if (board[m0][n1] != '.'){ s.insert(board[m0][n1]); }
                    }
                    if (m1 > m){
                        if (board[m1][n0] != '.'){ s.insert(board[m1][n0]); }
                        if (board[m1][n1] != '.'){ s.insert(board[m1][n1]); }
                    }
                    if (s.count(board[m][n])){ return false; }
                    s.clear();
                }
            }
        }
        return true;
    }
};

虽然程序执行效果还不错，运行时间和空间均超过了百分之80多的解答，但是代码比较复杂，可读性差，下面是别人家的改进解法。

扫描二维码关注公众号，回复： 9831885 查看本文章

别人家的解法：用算法代替了多个if定义

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        vector<unordered_set<int>> row(9);
        vector<unordered_set<int>> col(9);
        vector<unordered_set<int>> block(9);
        for(int i = 0; i < 9; ++ i){
            for(int j = 0; j < 9; ++ j){
                int bindex =  (i / 3)* 3 + j / 3;
                char cur = board[i][j];
                if(cur == '.')  continue;
                if(row[i].count(cur) || col[j].count(cur) || block[bindex].count(cur))  return false;
                row[i].insert(cur);
                col[j].insert(cur);
                block[bindex].insert(cur);
            }
        }
        return true;
    }
};

可以看到，将判断合三为一，大大减少了代码量。特别注意以下两点：

1. set、hash等容器都可以定义容器数组，此操作省去了连续定义多个容器的问题。
2. 注意这个代码：bindex = (i / 3)* 3 + j / 3，它的功能是定位到元素所在的9格块，此思想解决了我的代码中定义了一堆if的问题。