LeetCode 51. N-Queens
LeetCode题解专栏：LeetCode题解
我做的所有的LeetCode的题目都放在这个专栏里，大部分题目Java和Python的解法都有。
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.
Example:
Input: 4
Output: [
[".Q..", // Solution 1
"...Q",
"Q...",
"..Q."],
["..Q.", // Solution 2
"Q...",
"...Q",
".Q.."]
]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.
这道题是经典的N皇后问题，可以用回溯法解决
注意：Python和Java的解法是一样的，我看不懂C++的解法
大佬的python解法如下：
class Solution:
def solveNQueens(self, n: int) -> List[List[str]]:
def DFS(queens, xy_dif, xy_sum):
p = len(queens) # p is the index of row
if p == n:
result.append(queens)
return None
for q in range(n): # q is the index of col
# queens stores those used cols, for example, [0,2,4,1] means these cols have been used
# xy_dif is the diagonal 1
# xy_sum is the diagonal 2
if q not in queens and p - q not in xy_dif and p + q not in xy_sum:
DFS(queens + [q], xy_dif + [p - q], xy_sum + [p + q])
result = []
DFS([], [], [])
return [["." * i + "Q" + "." * (n - i - 1) for i in sol] for sol in result]
大佬的java解法如下：
//use backtracking algorithm to try out the positions
//for every level, there are n options to place. try them as in a small for loop
//from that level, go down, for each path, choose the next level node. check whether it's valid or not. if valid, then go on backtracking. (use a tmplist to store location info)
//if not valid, return.
//return condition is that it reached the final level and all queens placed
//question: what does it mean that two queens don't attach each other?
//怎么来看是否valid, 如何标记 每当加一个数字进去，相应的三个方向用数组来记录 mark下
//One puts a queen on the board and that immediately excludes one column, one row and two diagonals for the further queens placement.
//go back if not valid
class Solution {
//boolean array to track whether it can be placed or not.
//when one queen is placed on board, board[level][i], its column, its two diagonals are occupied.
boolean[] col;
boolean[] D45;
boolean[] D135; //这个D135的好难想
public List<List<String>> solveNQueens(int n) {
//create a grid and fill it.
char[][] board = new char[n][n];
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
board[i][j] = '.';
}
}
col = new boolean[n];
D45 = new boolean[2 * n];
D135 = new boolean[2 * n]; //default to false. meaning can place. true means occupied/can't place.
List<List<String>> res = new ArrayList<>();
if(n <= 0) return res;
backtrack(board, res, 0, n);
return res;
}
public void backtrack(char[][] board, List<List<String>> res, int level, int n) {
//if(level > n) return;
if(level == n) {
addSolution(res, board);
return;
}
for(int i = 0; i < n; i++) {
//if(!canPlace(board[level][i])) return;
if(col[i] || D45[level + i] || D135[i - level + n - 1]) continue; //确实是应该continue, return直接出去了, 这在for loop里面呢
//45度角的是 row + col.
//135是col - row.
//otherwise, continue
board[level][i] = 'Q';
col[i] = true;
D45[level + i] = true;
D135[i - level + n - 1] = true;
backtrack(board, res, level + 1, n);
board[level][i] = '.'; //复原 unmark for other routes.
col[i] = false;
D45[level + i] = false;
D135[i - level + n - 1] = false;
}
}
public void addSolution(List<List<String>> res, char[][]board) {
List<String> tmp = new ArrayList<>();
for(char[] level : board) {
tmp.add(String.valueOf(level)); //charArray to string. String.valueOf()
}
res.add(tmp);
}
}
大佬的c++解法如下：
class Solution {
public:
vector<vector<string>> solveNQueens(int n) {
vector<vector<string>> res;
vector<string> solution(n,std::string(n,'.'));
solve(res,solution,n,0,0,0);
return res;
}
void solve(vector<vector<string>> &res,vector<string> &solution,int n,int row, int ld ,int rd){
if(((1<<n)-1)==row){
res.push_back(solution);
return;
}
int D= ((1<<n)-1)&(~(row|rd|ld));
while(D){
int bit=D&(~D+1);
solution[rowcount(row)][n-colcount(bit)]='Q';
solve(res,solution,n,row|bit,(ld|bit)<<1,(rd|bit)>>1);
solution[rowcount(row)][n-colcount(bit)]='.';
D=D^bit;
}
}
int rowcount(int row){
int cnt=0;
while(row){
cnt+=row&0x1;
row>>=1;
}
return cnt;
}
int colcount(int col){
int cnt=0;
while(col){
cnt++;
col>>=1;
}
return cnt;
}
};
正常的C++解法如下：
class Solution {
public:
std::vector<std::vector<std::string> > solveNQueens(int n) {
std::vector<std::vector<std::string> > res;
std::vector<std::string> nQueens(n, std::string(n, '.'));
std::vector<int> flag_col(n, 1), flag_45(2 * n - 1, 1), flag_135(2 * n - 1, 1);
solveNQueens(res, nQueens, flag_col, flag_45, flag_135, 0, n);
return res;
}
private:
void solveNQueens(std::vector<std::vector<std::string> > &res, std::vector<std::string> &nQueens, std::vector<int> &flag_col, std::vector<int> &flag_45, std::vector<int> &flag_135, int row, int &n) {
if (row == n) {
res.push_back(nQueens);
return;
}
for (int col = 0; col != n; ++col)
if (flag_col[col] && flag_45[row + col] && flag_135[n - 1 + col - row]) {
flag_col[col] = flag_45[row + col] = flag_135[n - 1 + col - row] = 0;
nQueens[row][col] = 'Q';
solveNQueens(res, nQueens, flag_col, flag_45, flag_135, row + 1, n);
nQueens[row][col] = '.';
flag_col[col] = flag_45[row + col] = flag_135[n - 1 + col - row] = 1;
}
}
};