本文共 1251 字，大约阅读时间需要 4 分钟。

Determine if a Sudoku is valid, according to: .

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.

题解：验证矩阵的每一行是否有重复元素，每一列是否有重复元素，每一3X3的小矩形是否有重复元素，用visited数组来保存是已经存在的元素。

code：

public boolean isValidSudoku(char[][] board) {				boolean[] visited = new boolean[9];		//用来遍历每一行		for(int i=0; i<9; i++){			Arrays.fill(visited, false);			for(int j=0; j<9; j++){				if(!process(visited, board[i][j])){					return false;				}			}		}		//用来遍历每一列		for(int i=0; i<9; i++){			Arrays.fill(visited, false);			for(int j=0; j<9;j++){				if(!process(visited,board[j][i])){					return false;				}			}		}		//遍历小矩阵		for(int i=0; i<9; i+=3){						for(int j=0; j<9; j+=3){				Arrays.fill(visited, false);				for(int k=0; k<9; k++){					if(!process(visited, board[i+k/3][j+k%3])){						return false;					}				}			}		}		return true;	}	//判断是否为.和元素是否已经访问	public boolean process(boolean [] visited,char digit){				if(digit=='.'){			return true;		}		int num = digit - '0';		if(num< 1 || num>9 || visited[num-1]){			return false;		}		visited[num-1] = true;		return true;	}

转载地址：https://blog.csdn.net/zxdfc/article/details/48678347 如侵犯您的版权，请留言回复原文章的地址，我们会给您删除此文章，给您带来不便请您谅解！