首先,对于皇后的那个递归方法,我有三个变量分别表示1.一个int值,表示递归到当前的level,当前的哪几个col被占领了,例如11011就表示我们下一个Q只能放在第三个(从第一个位置开始数)位置上了;2.一个hash table, 表示左对角线到目前为止,是否有Queen占领过了;2.一个hash table,表示右对角线是否有Queen占领过了。
void recursive(int n,boolean[] column, boolean[] row, Set<Integer> leftSlash, Set<Integer> rightSlash, int[][] queen, List<List<String>> res) { if(n == 0){ generator(queen, res); return; } for(int i=0;i<column.length;i++){ int x = i; int y = column.length - n; //...
}private:voidtotalN(intn,intk){//放置第k个皇后if(k>n) sum++;else{for(inti=1;i<=n;i++){ board[k]=i;if(isok(k)) totalN(n,k+1); } } }boolisok(intk){//check whether the ktn Queen can be put downfor(inti=1;i<k;i++){if(board[i]==board[k]||abs(i-k)==abs(...
# 判断是否存在同对角线forj,idxinenumerate(queen):#len(queen)表示当前是第几个皇后ifabs(len(queen)-j)==abs(idx-i):flag=Truebreakifflag:continue# 合法则放入i列 queen.append(i)self.dfs(n,queen,ret)queen.pop()deftransform(self,n,ret):res=[]# 根据每个皇后摆放的列号还原棋盘forarrinret:...
class NQueens: """Generate all valid solutions for the n queens puzzle""" def __init__(self, size): # Store the puzzle (problem) size and the number of valid solutions self.size = size self.solutions = 0 self.solve() def solve(self): """Solve the n queens puzzle and print the...
for j, idx in enumerate(queen): # len(queen)表示当前是第几个皇后 if abs(len(queen) - j) == abs(idx - i): flag = True break if flag: continue # 合法则放入i列 queen.append(i) self.dfs(n, queen, ret) queen.pop() def transform(self, n, ret): ...
N-Queens II 题意: Follow up for N-Queens problem. Now, instead outputting board configurations, return the total number of distinct solutions. 思路: n 皇后问题,算可能的方案数。 主要的简单想法是对每一行处理。处理的时候尝试在每一列放一个皇后,仅仅要不冲突就向下递归,不断计算合法方案数。
N Queen II Deion: Follow up for N-Queens problem. Now, instead outputting board configurations, return the total number of distinct solutions. Input: 5 Output: 10 Day 1465 答案揭晓 DS Interview Question & Answer What is cross-validation? How to do it right?
(2 rows) N Queen II Deion: Follow up for N-Queens problem. Now, instead outputting board configurations, return the total number of distinct solutions. Input: 5 Output: 10 N Queens 的follow up, 用一个全局的计数器计算组合就行
Then-queens puzzle is the problem of placingnqueens on ann×nchessboard such that no two queens attack each other. n皇后问题是在n * n的棋盘上放置n个皇后,保证两两之间不会相互攻击。 国际象棋中的皇后比较牛逼,能攻击同一行、同一列、同一斜线上的棋子 ...