八皇后问题和数独问题
八皇后问题和数独问题是经典dfs+回溯的问题，其中还可以涉及hash，是非常好的练习题目。我们以leetcode中的四道题目为例。
Valid Sudoku
Total Accepted: 40598 Total Submissions: 149092Determine if a Sudoku is valid
The Sudoku board could be partially filled, where empty cells are filled with the character '.'.
A partially filled sudoku which is valid.
Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.
简单的判断数独表是否有解，只需要穷举每个元素判断其所在行、列以及小九宫格内没有和它相同的元素，为此我们可以建立hash表。
class Solution {
public:
bool isValidSudoku(vector<vector<char>>& board) {
int showed_col[9][9]={0}, showed_row[9][9]={0}, showed_sub[9][9]={0};
for(int i = 0 ; i < board.size();i++){
for(int j = 0; j < board[i].size(); j++){
int num = board[i][j] - '0' -1;
int sub = i/3*3+j/3;
if(board[i][j] !='.'){
if(showed_col[i][num] || showed_row[num][j] || showed_sub[sub][num])
return false;
showed_col[i][num] = 1;
showed_row[num][j] = 1;
showed_sub[sub][num] = 1;
}
}
}
return true;
}
};
Sudoku Solver
Total Accepted: 31158 Total Submissions: 142451
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.
求出数独的解，这里有个前提就是数独的解是唯一的，通过dfs+回溯解决。一共有两个核心函数，一个是回溯函数，一个是判断函数。
class Solution {
public:
void solveSudoku(vector<vector<char>>& board) {
solve(board,0);
}
bool isValidFill(vector<vector<char>> &board, int i,int j, char fill) {
for(int k = 0; k < 9; k++) {
if(board[i][k] == fill) return false;
if(board[k][j] == fill) return false;
int r = i/3*3+j/3;
if(board[r/3*3+k/3][r%3*3+k%3] == fill) return false;
}
return true;
}
bool solve(vector<vector<char>>&board, int ind) {
if(ind == 81) return true;
int i = ind/9, j = ind%9;
if(board[i][j] !='.') return solve(board, ind+1);
else{
for(char f = '1'; f<='9'; f++) {
if(isValidFill(board,i,j,f)){
board[i][j] = f;
if(solve(board,ind+1)) return true;
board[i][j] = '.';
}
}
}
return false;
}
};
下面我们会看到，八皇后问题与数独问题的求解十分相似。
N-Queens
Total Accepted: 34322 Total Submissions: 129991
The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.
Given an integer n, return all distinct solutions to the n-queens puzzle.
Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space respectively.
For example,
There exist two distinct solutions to the 4-queens puzzle:
[ [".Q..", // Solution 1 "...Q", "Q...", "..Q."], ["..Q.", // Solution 2 "Q...", "...Q", ".Q.."] ]
class Solution {
public:
vector<vector<string>> solveNQueens(int n) {
vector<vector<string>> res;
vector<string> nQueens(n, string(n,'.'));
solveNQueens(res, nQueens, 0, n);
return res;
}
bool isValid(vector<string> &nQueens, int row,int col,int &n){
for(int i=0; i != row; i++) {
if(nQueens[i][col] == 'Q')
return false;
}
for(int i = row -1, j = col -1; i >=0 && j>=0;--i,--j){
if(nQueens[i][j] == 'Q')
return false;
}
for(int i = row -1, j = col+1;i>=0&&j>=0;--i,++j){
if(nQueens[i][j] == 'Q')
return false;
}
return true;
}
void solveNQueens(vector<vector<string>> &res,vector<string> &nQueens,int row, int &n){
if(row == n) {
res.push_back(nQueens);
return;
}
for(int col = 0; col!=n;col++){
if(isValid(nQueens, row, col, n)) {
nQueens[row][col] = 'Q';
solveNQueens(res, nQueens,row+1,n);
nQueens[row][col] = '.';
}
}
}
};
N-Queens II
Total Accepted: 29382 Total Submissions: 81802
Follow up for N-Queens problem.
Now, instead outputting board configurations, return the total number of distinct solutions.
class Solution {
public:
int totalNQueens(int n) {
int result = 0;
vector<int> A(n,-1);
NQueens(n,result,0,A);
return result;
}
bool isValid(vector<int> &A, int r){
for(int i = 0; i < r; i++) {
if((A[i] == A[r]) || (abs(A[i]-A[r]) == (r-i)))
return false;
}
return true;
}
void NQueens(int n, int &result, int cur, vector<int> &A) {
if(cur == n) {
result++;
return;
}else{
for(int i = 0; i < n; i++) {
A[cur] = i;
if(isValid(A,cur))
NQueens(n,result,cur+1,A);
}
}
}
};