判斷一個(gè) 9x9 的數(shù)獨(dú)是否有效。只需要根據(jù)以下規(guī)則，驗(yàn)證已經(jīng)填入的數(shù)字是否有效即可男旗。
數(shù)字 1-9 在每一行只能出現(xiàn)一次颅眶。
數(shù)字 1-9 在每一列只能出現(xiàn)一次。
數(shù)字 1-9 在每一個(gè)以粗實(shí)線分隔的 3x3 宮內(nèi)只能出現(xiàn)一次止吐。
數(shù)獨(dú)部分空格內(nèi)已填入了數(shù)字，空白格用 '.' 表示。
示例 1:
輸入:
[
["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"]
]
輸出: true
來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/valid-sudoku
解題思路
HashMap 的空間消耗大, 采用布爾二維數(shù)組
boolean[][] rows = new boolean[9][9]
boolean[][] cols = new boolean[9][9]
boolean[][] boxes = new boolean[9][9]
rows[i][num - 1]代表第 i 行的數(shù) num 是否出現(xiàn)過, 是則為 true
因?yàn)閿?shù)組下標(biāo)為0到長度 - 1, 所以用 num - 1 代表num
遍歷9 * 9的數(shù)獨(dú)時(shí), 依次判斷遍歷的數(shù)字在該行該列該九宮格是否出現(xiàn)過, 出現(xiàn)過立刻返回 false, 沒出現(xiàn)過則設(shè)為出現(xiàn)過 true
遍歷完之前沒返回結(jié)果則說明符合有效數(shù)獨(dú)
代碼
class Solution {
public boolean isValidSudoku(char[][] board) {
boolean[][] rows = new boolean[9][9];
boolean[][] cols = new boolean[9][9];
boolean[][] boxes = new boolean[9][9];
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
char numChar = board[i][j];
if (numChar != '.') {
int num = numChar - '0';
if (rows[i][num - 1]) {
return false;
} else {
rows[i][num - 1] = true;
}
if (cols[j][num - 1]) {
return false;
} else {
cols[j][num - 1] = true;
}
if (boxes[i / 3 * 3 + j / 3][num - 1]) {
return false;
} else {
boxes[i / 3 * 3 + j / 3][num - 1] = true;
}
}
}
}
return true;
}
}
