LeetCode37. Sudoku Solver
2016-12-10 14:50
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题意:解9*9数独
做法:回溯
对每一个还没填充的格子,尝试1~9这9个数字,如果是合法的,则继续填充下一个格子,否则回溯。
判断合法:只需要判断对应行、对应列、对应3*3,有没有矛盾。
设未被填充的格子个数为n,那么时间复杂度大约为O(9n+1)=O(9n)
在leetcode上用时76ms,本题在leetcode上有0ms的解法,
2ms解法
0ms解法
做法:回溯
对每一个还没填充的格子,尝试1~9这9个数字,如果是合法的,则继续填充下一个格子,否则回溯。
判断合法:只需要判断对应行、对应列、对应3*3,有没有矛盾。
设未被填充的格子个数为n,那么时间复杂度大约为O(9n+1)=O(9n)
在leetcode上用时76ms,本题在leetcode上有0ms的解法,
2ms解法
0ms解法
class Solution { public: void solveSudoku(vector<vector<char>>& board) { solve(board); } private: bool solve(vector<vector<char>>& board){ for(int i = 0; i < board.size(); ++i) for(int j = 0; j < board[0].size(); ++j){ if(board[i][j] == '.'){ for(int ch = '1'; ch <= '9'; ++ch) if(isValid(board, i, j, ch)){ board[i][j] = ch; if(solve(board)) return true; else board[i][j] = '.'; } return false; } } return true; } bool isValid(vector<vector<char>>& board, int row, int col, char ch) { for(int i = 0; i < board.size(); ++i){ int x = row / 3 * 3 + i / 3, y = col / 3 * 3 + i % 3; if(board[i][col] == ch || board[row][i] == ch || board[x][y] == ch) return false; } return true; } };
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