1. 解数独
37. 解数独
https://leetcode.cn/problems/sudoku-solver/
编写一个程序，通过填充空格来解决数独问题。
数独的解法需 遵循如下规则：
- 数字 1-9 在每一行只能出现一次。
- 数字 1-9 在每一列只能出现一次。
- 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。（请参考示例图）
数独部分空格内已填入了数字，空白格用 '.' 表示。
示例 1：
输入：board = [["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"]]
输出：[["5","3","4","6","7","8","9","1","2"],
["6","7","2","1","9","5","3","4","8"],
["1","9","8","3","4","2","5","6","7"],
["8","5","9","7","6","1","4","2","3"],
["4","2","6","8","5","3","7","9","1"],
["7","1","3","9","2","4","8","5","6"],
["9","6","1","5","3","7","2","8","4"],
["2","8","7","4","1","9","6","3","5"],
["3","4","5","2","8","6","1","7","9"]]
解释：输入的数独如上图所示，唯一有效的解决方案如下所示：
解题思路
这是一道hard。3个要求肯定需要三组标记进行if判断。而且很明显这是一道穷举的带回溯的题，需要使用到回溯算法，也就是深度优先。
这是一个固定9x9大小的框，为了满足三个要求，所以需要先构建三个标记set或者list。Set含有一个求交集的API，所以Set比较合适。所以有一个colSet[]记录每一列的可选元素，rowset[]记录每一行的可选元素，boxSet[]每一个框的可选元素。
每次都从交集里面取出一个数字，然后进入下一个迭代。当到达8，9的时候就结束。
思路有部分的bug：
当row为9的时候结束返回true。
代码
class Solution {HashSet<Character>[] setCol = new HashSet[9];HashSet<Character>[] setRow = new HashSet[9];HashSet<Character>[] setBox = new HashSet[9];public void solveSudoku(char[][] board) {initializeSets(board);backTrack(board, 0, 0);}private void initializeSets(char[][] board) {HashSet<Character> full = new HashSet<>(Arrays.asList('1', '2', '3', '4', '5', '6', '7', '8', '9'));for (int i = 0; i < 9; i++) {setRow[i] = new HashSet<>(full);setCol[i] = new HashSet<>(full);setBox[i] = new HashSet<>(full);}for (int i = 0; i < 9; i++) {for (int j = 0; j < 9; j++) {char num = board[i][j];if (num != '.') {setRow[i].remove(num);setCol[j].remove(num);setBox[i / 3 * 3 + j / 3].remove(num);}}}}private boolean backTrack(char[][] board, int i, int j) {if (i == 9)return true;if (j == 9)return backTrack(board, i + 1, 0);if (board[i][j] != '.') {return backTrack(board, i, j + 1);}HashSet<Character> set = new HashSet<>(setRow[i]);set.retainAll(setCol[j]);set.retainAll(setBox[i / 3 * 3 + j / 3]);for (char num : set) {board[i][j] = num;setRow[i].remove(num);setCol[j].remove(num);setBox[i / 3 * 3 + j / 3].remove(num);if (backTrack(board, i, j + 1)) {return true;}board[i][j] = '.';setRow[i].add(num);setCol[j].add(num);setBox[i / 3 * 3 + j / 3].add(num);}return false;}
}
