2014
11-19

LeetCode-Sudoku Solver[DFS]

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.

// LeetCode, Sudoku Solver
// 时间复杂度O(9^4)，空间复杂度O(1)
class Solution {
public:
bool solveSudoku(vector<vector<char> > &board) {
for (int i = 0; i < 9; ++i)
for (int j = 0; j < 9; ++j) {
if (board[i][j] == '.') {
for (int k = 0; k < 9; ++k) {
board[i][j] = '1' + k;
if (isValid(board, i, j) && solveSudoku(board))
return true;
board[i][j] = '.';
}
return false;
}
}
return true;
}
private:
// 检查 (x, y) 是否合法
bool isValid(const vector<vector<char> > &board, int x, int y) {
int i, j;
for (i = 0; i < 9; i++) // 检查 y 列
if (i != x && board[i][y] == board[x][y])
return false;
for (j = 0; j < 9; j++) // 检查 x 行
if (j != y && board[x][j] == board[x][y])
return false;
for (i = 3 * (x / 3); i < 3 * (x / 3 + 1); i++)
for (j = 3 * (y / 3); j < 3 * (y / 3 + 1); j++)
if ((i != x || j != y) && board[i][j] == board[x][y])
return false;
return true;
}
};


1. 嗯 分析得很到位，确实用模板编程能让面试官对你的印象更好。在设置辅助栈的时候可以这样：push时，比较要push的elem和辅助栈的栈顶，elem<=min.top()，则min.push(elem).否则只要push（elem）就好。在pop的时候，比较stack.top()与min.top(),if(stack.top()<=min.top()),则{stack.pop();min.pop();}，否则{stack.pop();}.

2. 因为是要把从字符串s的start位到当前位在hash中重置，修改提交后能accept，但是不修改居然也能accept