题目： 

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解析： 

部分决策树： 

 

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代码设计&剪枝&回溯： 

 

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代码： 

class Solution {private boolean[][] row, col;private boolean[][][] gird; public void solveSudoku(char[][] board) {//下标->数字；0->1, 1->2row = new boolean[9][10];col = new boolean[9][10];gird = new boolean[3][3][10];//初始化上面的标记数组for(int i = 0; i < 9; i++)for(int j = 0; j < 9; j++){int num = board[i][j]-'0';if(board[i][j] != '.'){row[i][num] = col[j][num] = gird[i/3][j/3][num] = true;}}dfs(board);}private boolean dfs(char[][] board){for(int i = 0; i < 9; i++){for(int j = 0; j < 9; j++){if(board[i][j] == '.'){for(int num = 1; num <= 9; num++){//剪枝写法if(!row[i][num] && !col[j][num] && !gird[i/3][j/3][num]){board[i][j] = (char)('0' + num);row[i][num] = col[j][num] = gird[i/3][j/3][num] = true;//填数字往下遍历时候可能会出现 “某一行无数可以填”if(dfs(board) == true) return true;//回溯board[i][j] = '.';row[i][num] = col[j][num] = gird[i/3][j/3][num] = false;}}//一整行都没有返回时(已经试过9个数)，也是出现“某一行无数可以填”return false;}}}//上面没有返回代表，前面的dfs已经全部填完return true;}
}