LeetCode 51. N-Queens--回溯法 pyhon,java,c++解法

题目地址：

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 rowif p == n:result.append(queens)return Nonefor 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 2if 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 validclass 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, continueboard[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;}}
};

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LeetCode 51. N-Queens--回溯法 pyhon,java,c++解法

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