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【LeetCode从零单排】No36 Valid Sudoku

时间:2015-02-13 18:38:44      阅读:114      评论:0      收藏:0      [点我收藏+]

标签:leetcode   java   

题目

      判断数独是否成立的一道题,看的是某大神的答案,写的太漂亮了。

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 ‘.‘.

技术分享

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.


代码

public class Solution {
	public boolean isValidSudoku(char[][] board) {
	    int count = 9;
	    int blockWidth = 3;
	    for (int i=0; i<count; i++) {
	        boolean[] rowExist = new boolean[count+1];
	        boolean[] colExist = new boolean[count+1];
	        boolean[] matrixExist = new boolean[count+1];
	        for (int j=0; j<count; j++) {
	            int rowNum = board[i][j] == ‘.‘ ? -1 : board[i][j] - ‘0‘;
	            int colNum = board[j][i] == ‘.‘ ? -1 : board[j][i] - ‘0‘;
	            int mtxRowIdx = 3*(i/3);
	            int mtxColIdx = 3*(i%3);
	            int matrixNum = board[mtxRowIdx + j/3][mtxColIdx + j%3] == ‘.‘ ? 
	                                -1 : board[mtxRowIdx + j/3][mtxColIdx + j%3] - ‘0‘;
	            if (rowNum > 0 && rowExist[rowNum] ||
	                colNum > 0 && colExist[colNum] || 
	                matrixNum > 0 && matrixExist[matrixNum]) {
	                return false;        
	            }
	            if (rowNum > 0)
	                rowExist[rowNum] = true;
	            if (colNum > 0)
	                colExist[colNum] = true;
	            if (matrixNum > 0)
	                matrixExist[matrixNum] = true;
	        }
	    }
	    return true;
	}}



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* 本文来自博客  “李博Garvin“

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【LeetCode从零单排】No36 Valid Sudoku

标签:leetcode   java   

原文地址:http://blog.csdn.net/buptgshengod/article/details/43795705

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