"Leetcode" N-queens II

Test instructions

Follow up for n-queens problem.

Now, instead outputting board configurations and return the total number of distinct solutions.

Ideas:

N Queen question, count the number of possible scenarios. The basic simple idea is to deal with each line, try to put a queen in each column, as long as there is no conflict, recursive downward, constantly calculate the number of legitimate programs.

Code:

C++

Class Solution {Public:int totalnqueens (int n) {/* Initialize vector variable, number I represents the queen of line I is in which column */Vector<int&gt ;        Chess (n,-1);        int ans = 0;        /* Solve the problem */Solvequeen (0,n,chess.begin (), ans);    return ans; } void Solvequeen (int r,int n,vector<int>::iterator chess,int &ans) {/*r equals n when each row has Queen */I            F (r = = N) {ans++;        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 row            */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; }};

Python:

Class solution:    # @return An integer    def totalnqueens (self, N):        Self.array = [0 for I in range (0,n)] self        . Ans = 0        self.slovequeen (0,n)        return Self.ans    def slovequeen (self,r,n):        if r = = N:            Self.ans + = 1            return for        I in range (0,n):            self.array[r] = i            if Self.check (r,n):                self.slovequeen (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 II