LeetCode Valid Sudoku

最新推荐文章于 2024-01-06 14:41:27 发布

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最新推荐文章于 2024-01-06 14:41:27 发布

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分类专栏： LeetCode 文章标签： leetcode hashmap matrix 模拟

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本文介绍了一种通过遍历数独的每一行、每一列和每一个区块来验证其有效性的算法，确保0-9数字在每个区块内只出现一次。

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Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

A partially filled sudoku which is valid.

Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

对每行、每列、每块都进行一次遍历，保证0-9只能出现一次，采用int num[10]进行标记

char转int：char-'0'即可

java中类似memset方法是Arrays.fill()方法

public class Solution {
    public boolean isValidSudoku(char[][] board) {
        int[] num=new int[10];
        for(int i=0;i<9;i++){   //每行
        	Arrays.fill(num, 0);  //赋值0
        	for(int j=0;j<9;j++){
        		if(board[i][j]!='.'&&num[board[i][j]-'0']==1)
        			return false;
        		if(board[i][j]!='.')
        			num[board[i][j]-'0']=1;
        	}
        }
        for(int j=0;j<9;j++){   //每列
        	Arrays.fill(num, 0);
        	for(int i=0;i<9;i++){
        		if(board[i][j]!='.'&&num[board[i][j]-'0']==1)
        			return false;
        		if(board[i][j]!='.')
        			num[board[i][j]-'0']=1;
        	}
        }
        for(int j=0;j<9;j+=3){    //每块
        	for(int i=0;i<9;i+=3){
        		Arrays.fill(num, 0);
        		for(int k=0;k<3;k++){
        			for(int t=0;t<3;t++){
        				if(board[j+k][i+t]!='.'&&num[board[j+k][i+t]-'0']==1)
        					return false;
        				if(board[j+k][i+t]!='.')
        					num[board[j+k][i+t]-'0']=1;
        			}
        		}
        	}
        }
        return true;
    }
}