Test instructions

The n-queens Puzzle is the problem of placing N Queens on a nxn chessboard such that No, Queens attack.

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 Indi Cate a queen and an empty space respectively.

For example,

There exist-distinct solutions to the 4-queens puzzle:

[ [". Q.. ", //solution 1 " ... Q ", " Q ... ",". . Q. "], ["]. Q. ", //Solution 2 " Q ... ", " ... Q ", ". Q.. "]

Ideas

N Queen problem, but to output each solution. Just give each line a recursive go.

Code

C++:

Class Solution {public:vector<vector<string> > Solvenqueens (int n) {/* Initialize vector variable, number of I represents line I The Queen in which column */vector<int> chess (n,-1); /* Save Results */vector< vector<string> > ans; /* Solve the problem */Solvequeen (0,n,chess.begin (), ans); return ans; } void Solvequeen (int r,int n,vector<int>::iterator chess,vector< vector<string> > &ans) { /*r equals n When each row has a queen * * if (r = = N) {/*solution is used to hold a legal solution */vector<string> Solu tion; for (int i = 0;i < N;++i) {Solution.push_back (getrowinstring (n,*)); } ans.push_back (solution); Return }/* To see which column of the current row can be put Queen */for (int i = 0;i < N;++i) {* (chess+r) = i; /* Check legality */if (check (chess,r,n)) {/* down recursion */Solvequeen (R+1,n,chess,ans); } }/* Check for conflicts */bool Check (Vector<int>::iterator chess,int r,int N) {/* for each previous line */For (int i = 0;i < R;++i) {/* Calculates the distance between two columns */int dis = ABS (* (CHESS+R)-* (Chess+i)); /* dis = 0 in the same column, dis = r-1 constitutes isosceles triangle, i.e. diagonal */if (dis = = 0 | | dis = = r-i) return false; } return true; }/* Constructs n-length string getrowinstring (int n,int col) in Col for Queen, {string str (n, '. '); Str.replace (col,1, "Q"); return str; }};

Python:

Class solution: # @return An integer def solvenqueens (self, N): Self.array = [0 for I in range (0,n)] self . Ans = [] self.solve (0,n) return Self.ans def solve (self,r,n): if r = = N: solution = [] for i in RA Nge (0,n): row = ['. ' For x in range (0,n)] row[self.array[i]] = ' Q ' solution.append (". Join (Row)) Self.ans.append (solution) del solution return for I in range (0,n): self.array[r] = i if Self.check (r,n): self.solve (r+1,n) def check (self,r,n): For I in range (0,r): dis = ABS ( SELF.ARRAY[R]-self.array[i]) if dis = = 0 or dis = = r-i: return False return True

"Leetcode" N-queens