[leetcode]Valid Sudoku

简单题。先把横的竖的扫一下，然后按照每个3*3的小格扫一遍。

import java.util.HashSet;

public class Solution {
    public boolean isValidSudoku(char[][] board) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if (board.length == 0) return false;
        if (board.length % 3 != 0) return false;
        if (board[0].length != board.length) return false;
        // check length of each row
        for (int i = 0; i < board.length; i++) {
            for (int j = 1; j < board[0].length; j++) {
        		if (board[j].length != board[0].length) return false;
        	}
        }
        
        int len = board.length;
        for (int i = 0; i < len; i++) {
        	HashSet<Character> set1 = new HashSet<Character>();
        	HashSet<Character> set2 = new HashSet<Character>();
        	for (int j = 0; j < len; j++) {
        		char c1 = board[i][j];
        		if (c1 == '.') {}
        		else if (c1 >= '1' && c1 <= '9') {
        			if (set1.contains(c1)) {
        				return false;
        			}
        			set1.add(c1);
        		}
        		else { return false; }
        		char c2 = board[j][i];
        		if (c2 == '.') {}
        		else if (c2 >= '1' && c2 <= '9') {
        			if (set2.contains(c2)) {
        				return false;
        			}
        			set2.add(c2);
        		}
        		else { return false; }
        	}
        }
        
        int len_3 = len / 3;
        
        for (int i = 0; i < len_3; i++) {
        	for (int j = 0; j < len_3; j++) {
        		HashSet<Character> set = new HashSet<Character>();
        		for (int x = 0; x < 3; x++) {
        			for (int y = 0; y < 3; y++) {
        				
        				char c = board[i*3+x][j*3+y];
                		if (c == '.') {}
                		else if (c >= '1' && c <= '9') {
                			if (set.contains(c)) {
                				return false;
                			}
                			set.add(c);
                		}
                		else { return false; }
        			}
        		}
        	}
        }
        return true;
    }
}

看了一下参考，其实一边扫描就可以了。任何一个格子扫完都能把三个判断条件同时往前推进。参考：http://discuss.leetcode.com/questions/215/valid-sudoku

class Solution {
public:
    bool isValidSudoku(vector<vector<char> > &board) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<bool> > rows(9, vector<bool>(9, false));
        vector<vector<bool> > cols(9, vector<bool>(9, false));
        vector<vector<bool> > blocks(9, vector<bool>(9, false));

        for (int i = 0; i < 9; ++i) {
            for (int j = 0; j < 9; ++j) {
                if (board[i][j] == '.') continue;
                int c = board[i][j] - '1';
                if (rows[i][c] || cols[j][c] || blocks[i - i % 3 + j / 3][c])
                    return false;
                rows[i][c] = cols[j][c] = blocks[i - i % 3 + j / 3][c] = true;
            }
        }
        return true;
    }
};

第二刷：

Annie的做法也是只扫一遍，而且用了bit来存，省空间。