LeetCode刷题之有效数独

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题目描述

请你判断一个 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。（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。

注意：

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。

示例 1：

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：true

思路

1. 数独验证的问题可以将其转化为精确覆盖问题
2. 将题目提供的三个要求转化为相互独立的要求

• 如果定义第i行，第j列的数字：num
  • 对于要求一，可以划分为，数字num出现在第i行，那么也就是一共81个条件
  • 要求二和要求三与要求一同理，共81*2=162个

于是我们使用一个长度为243的数组来记录一个条件是否满足。并通过行列数来计算得出该条件的下标

• 要求一：first = i+(num-1)*9
• 要求二：second = 81+j+(num-1)*9
• ``要求三：third = 162+(i/3)*3+j/3+(num-1)*9`

遍历二维数组，对于数字我们通过上面的计算式的结果作为数组下标，然后数值加一,最后遍历用于记录的数组，如果出现某个位置的值大于等于2.则说明该要求被重复满足了，则数独不合理

class Solution {
    public boolean isValidSudoku(char[][] board) {
        int[] judge = new int[243];
        for(int i=0; i<9; i++) {
            for(int j=0; j<9; j++) {
                if(board[i][j]!='.') {
                    int num = (int)board[i][j]-48;
                    int first = i+(num-1)*9;
                    int second = 81+j+(num-1)*9;
                    int third = 162+(i/3)*3+j/3+(num-1)*9;
                    judge[first]++;
                    judge[second]++;
                    judge[third]++;
                }
            }
        }
        for(int i=0; i<243; i++){
            if(judge[i]>=2){
                return false;
            }
        }
        return true;
    }
}