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wangpeng
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feat(MEDIUM): add _36_isValidSudoku
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package pp.arithmetic.leetcode;
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import java.util.HashMap;
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/**
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* Created by wangpeng on 2019-05-14.
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* 36. 有效的数独
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*
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* 判断一个 9x9 的数独是否有效。只需要根据以下规则,验证已经填入的数字是否有效即可。
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*
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* 数字 1-9 在每一行只能出现一次。
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* 数字 1-9 在每一列只能出现一次。
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* 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。
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*
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* <image src="https://upload.wikimedia.org/wikipedia/commons/thumb/f/ff/Sudoku-by-L2G-20050714.svg/250px-Sudoku-by-L2G-20050714.svg.png" />
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* 上图是一个部分填充的有效的数独。
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*
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* 数独部分空格内已填入了数字,空白格用 '.' 表示。
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*
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* 示例 1:
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*
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* 输入:
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* [
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* ["5","3",".",".","7",".",".",".","."],
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* ["6",".",".","1","9","5",".",".","."],
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* [".","9","8",".",".",".",".","6","."],
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* ["8",".",".",".","6",".",".",".","3"],
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* ["4",".",".","8",".","3",".",".","1"],
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* ["7",".",".",".","2",".",".",".","6"],
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* [".","6",".",".",".",".","2","8","."],
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* [".",".",".","4","1","9",".",".","5"],
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* [".",".",".",".","8",".",".","7","9"]
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* ]
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* 输出: true
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* 示例 2:
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*
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* 输入:
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* [
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* ["8","3",".",".","7",".",".",".","."],
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* ["6",".",".","1","9","5",".",".","."],
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* [".","9","8",".",".",".",".","6","."],
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* ["8",".",".",".","6",".",".",".","3"],
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* ["4",".",".","8",".","3",".",".","1"],
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* ["7",".",".",".","2",".",".",".","6"],
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* [".","6",".",".",".",".","2","8","."],
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* [".",".",".","4","1","9",".",".","5"],
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* [".",".",".",".","8",".",".","7","9"]
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* ]
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* 输出: false
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* 解释: 除了第一行的第一个数字从 5 改为 8 以外,空格内其他数字均与 示例1 相同。
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* 但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。
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* 说明:
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*
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* 一个有效的数独(部分已被填充)不一定是可解的。
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* 只需要根据以上规则,验证已经填入的数字是否有效即可。
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* 给定数独序列只包含数字 1-9 和字符 '.' 。
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* 给定数独永远是 9x9 形式的。
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*/
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public class _36_isValidSudoku {
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public static void main(String[] args) {
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char[][] board = new char[][]{
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{'5', '3', '.', '.', '7', '.', '.', '.', '.'},
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{'6', '.', '.', '1', '9', '5', '.', '.', '.'},
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{'.', '9', '8', '.', '.', '.', '.', '6', '.'},
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{'8', '.', '.', '.', '6', '.', '.', '.', '3'},
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{'4', '.', '.', '8', '.', '3', '.', '.', '1'},
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{'7', '.', '.', '.', '2', '.', '.', '.', '6'},
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{'.', '6', '.', '.', '.', '.', '2', '8', '.'},
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{'.', '.', '.', '4', '1', '9', '.', '.', '5'},
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{'.', '.', '.', '.', '8', '.', '.', '7', '9'}
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};
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_36_isValidSudoku isValidSudoku = new _36_isValidSudoku();
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System.out.println(isValidSudoku.isValidSudoku(board));
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}
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/**
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* 解题思路:
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* 首先必须理解有效数独的定义
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* 要知道是否满足数独的要求,肯定所有的点都得遍历到,所以得双重遍历
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* 在双重遍历中得匹配之前的遍历结果,看是否有重复的数字,考虑用hashmap报错遍历结果
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* hashmap中存储的key是有三种:行的index+数字、列的index+数字、3*3宫格的index+数字
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*
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* 提交结果:
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* 执行用时 : 16 ms, 在Valid Sudoku的Java提交中击败了62.19% 的用户
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* 内存消耗 : 42.4 MB, 在Valid Sudoku的Java提交中击败了81.42% 的用户
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* 结果并不是很出色,看了其他优秀解法,发现他们存储是固定大小三个数组,
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* 仔细想想也是,HashMap虽然查找是O(1),但是扩容的时候是有时间消耗的,对于这种固定大小的可以考虑用数组进行优化
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* 官方解题,是分了三个HashMap,估计也是为了减少扩容的时间消耗吧
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*
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* @param board
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* @return
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*/
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public boolean isValidSudoku(char[][] board) {
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HashMap<String, Boolean> map = new HashMap<>();
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for (int i = 0; i < board.length; i++) {
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for (int j = 0; j < board[i].length; j++) {
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char num = board[i][j];
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if (num == '.') continue;
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String rowKey = i + "row" + num;
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String colKey = j + "col" + num;
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int groupIndex = i / 3 + j / 3 * 3;
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String groupKey = groupIndex + "group" + num;
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//寻找是否有重复的数字
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if (map.getOrDefault(rowKey, false)
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|| map.getOrDefault(colKey, false)
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|| map.getOrDefault(groupKey, false)) {
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return false;
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}
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//更新遍历记录
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map.put(rowKey, true);
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map.put(colKey, true);
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map.put(groupKey, true);
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}
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}
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return true;
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}
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}

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