LeetCode 37. Sudoku Solver--回溯法

题目链接

37. Sudoku Solver

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...

...and its solution numbers marked in red.

解：其实和八皇后的问题是类似的，只是状态和判断冲突复杂了一些，首先要清楚数独的规则，每一行每一列是1～9，每个小九宫格是1～9，（注意对角线不一定是1～9！！），从左上角的位置开始向右向下填，遇到数字直接跳过，当在某个位置没有数字可填时，回溯，注意在回溯过程中要把已经填的数字都恢复到 ‘.’ 原状态，当填到右下角的位置，且那个位置是固定数字or可以找到一个不冲突的数字填入，说明找到了解，代码：

class Solution {
public:
	bool isConflict(int line, int col, char value, vector >& board) {
		for (int i = 0; i < 9; i++) {
			if (board[i][col] == value) return true;
			if (board[line][i] == value) return true;
		}
		int line_no = line/3;
		int col_no = col/3;
		for (int i = 0; i < 3; i++) 
			for (int j = 0; j < 3; j++) 
				if (board[line_no*3+i][col_no*3+j] == value) return true;
		return false;
	}

	bool solveOneStep(int line, int col, vector >& board) {
		if (board[line][col] != '.') {
			if (col == 8 && line == 8) return true;
			if (col == 8) return solveOneStep(line+1, 0, board);
			else return solveOneStep(line, col+1, board);
		}
		bool result = false;
		for (int i = 1; i <= 9; i++) {
			board[line][col] = '.';
			if (!isConflict(line, col, '0'+i, board)) {
				board[line][col] = '0'+i;
				if (line == 8 && col == 8) return true;
				if (col == 8) result = solveOneStep(line+1, 0, board);
				else result = solveOneStep(line, col+1, board);
				if (result) break;
			}
		}
		if (!result) board[line][col] = '.';
		return result;
	}

    void solveSudoku(vector >& board) {
        solveOneStep(0, 0, board);
    }
};