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解决数独
Write a program to solve a Sudoku puzzle by filling the empty cells.
Empty cells are indicated by the character ‘.‘.
You may assume that there will be only one unique solution.
A sudoku puzzle...
思路：
搜索加剪枝。
首先设立3个辅助二维数组：rows, columns, grids来维护目前的数独状态，rows[i][k]（1<=i, k<=9）表示第i行是否占用了数字k，同理colums[i][k]表示第i列是否占用了数字k，grid[i][k]表示第i个格子是否占用了数字k，
然后第一次读入二维数组，更新rows, columns和grids.
第二次读二维数组，对每个为‘.‘的元素根据rows,columns和grid来枚举可能的值。如果没有可以枚举的值，直接剪枝返回。
代码如下：
Runtime: 5 ms
static int columns[9][10]; static int rows[9][10]; static int grids[9][10]; int search(char *board[9], int startI, int startJ) { for(int i = startI; i < 9; ++i) { for(int j = i == startI ? startJ : 0; j < 9; ++j) { if (board[i][j] == ‘.‘) { for (int k = 1; k <= 9; ++k) { if (!rows[i][k] && !columns[j][k] && !grids[i/3 + j/3 *3][k]) { rows[i][k] = 1; columns[j][k] = 1; grids[i/3 + j/3 *3][k] = 1; if (search(board, i, j+1) == 1) { board[i][j] = k + ‘0‘; return 1; } rows[i][k] = 0; columns[j][k] = 0; grids[i/3 + j/3 *3][k] = 0; } } return 0; } } } return 1; } void solveSudoku(char * board[9]) { memset(columns, 0, sizeof(columns)); memset(rows, 0, sizeof(rows)); memset(grids, 0, sizeof(grids)); for(int i = 0; i < 9; ++i) { for(int j = 0; j < 9; ++j) { if (board[i][j] != ‘.‘) { rows[i][board[i][j] - ‘0‘] = 1; columns[j][board[i][j] - ‘0‘] = 1; int tmp = i/3 + j/3 * 3; grids[tmp][board[i][j] - ‘0‘] = 1; } } } search(board, 0, 0); }
leetcode problem 37 -- Sudoku Solver
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原文地址：http://www.cnblogs.com/lysuns/p/4454036.html
