题目:
解题思路:
回溯法解数独
类似人的思考方式去尝试,行,列,还有 3*3 的方格内数字是 1~9 不能重复。
我们尝试填充,如果发现重复了,那么擦除重新进行新一轮的尝试,直到把整个数组填充完成。
https://leetcode-cn.com/problems/sudoku-solver/solution/hui-su-fa-jie-shu-du-by-i_use_python/
代码:
class Solution {
public void solveSudoku(char[][] board) {
// 三个布尔数组 表明 行, 列, 还有 3*3 的方格的数字是否被使用过
boolean[][] rowUsed = new boolean[9][10];
boolean[][] colUsed = new boolean[9][10];
boolean[][][] boxUsed = new boolean[3][3][10];
// 初始化
for(int row = 0; row < board.length; row++){
for(int col = 0; col < board[0].length; col++) {
int num = board[row][col] - '0';
if(1 <= num && num <= 9){
rowUsed[row][num] = true;
colUsed[col][num] = true;
boxUsed[row/3][col/3][num] = true;
}
}
}
// 递归尝试填充数组
recusiveSolveSudoku(board, rowUsed, colUsed, boxUsed, 0, 0);
}
private boolean recusiveSolveSudoku(char[][]board, boolean[][]rowUsed, boolean[][]colUsed, boolean[][][]boxUsed, int row, int col){
// 边界校验, 如果已经填充完成, 返回true, 表示一切结束
if(col == board[0].length){
col = 0;
row++;
if(row == board.length){
return true;
}
}
// 是空则尝试填充, 否则跳过继续尝试填充下一个位置
if(board[row][col] == '.') {
// 尝试填充1~9
for(int num = 1; num <= 9; num++){
boolean canUsed = !(rowUsed[row][num] || colUsed[col][num] || boxUsed[row/3][col/3][num]);
if(canUsed){
rowUsed[row][num] = true;
colUsed[col][num] = true;
boxUsed[row/3][col/3][num] = true;
board[row][col] = (char)('0' + num);
if(recusiveSolveSudoku(board, rowUsed, colUsed, boxUsed, row, col + 1)){
return true;
}
board[row][col] = '.';
rowUsed[row][num] = false;
colUsed[col][num] = false;
boxUsed[row/3][col/3][num] = false;
}
}
} else {
return recusiveSolveSudoku(board, rowUsed, colUsed, boxUsed, row, col + 1);
}
return false;
}
}
来源:oschina
链接:https://my.oschina.net/u/4248053/blog/4415090