leetcode -- N-Queens I&II，经典回溯，再看

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https://leetcode.com/problems/n-queens/

https://leetcode.com/problems/n-queens-ii/

参考
递归回溯
http://www.cnblogs.com/zuoyuan/p/3747249.html
其中还可以加上board[depth]= -1

 board[depth]=i s='.'*n dfs(depth+1, valuelist+[s[:i]+'Q'+s[i+1:]]) board[depth]= -1 #可要可不要

class Solution:    # @return a list of lists of string    def solveNQueens(self, n):        def check(k, j):  # check if the kth queen can be put in column j!            for i in range(k):                if board[i]==j or abs(k-i)==abs(j - board[i]):#这里当前点为(k, j)以及(i, board[i])                #board[i]==j就是说不能列相同                #abs(k-i)==abs(j - board[i])，不能行差等于列差，即在对角线上                    return False            return True        def dfs(depth, valuelist):            if depth==n: res.append(valuelist); return            for i in range(n):                if check(depth,i):                     board[depth]=i                    s='.'*n                    dfs(depth+1, valuelist+[s[:i]+'Q'+s[i+1:]])        board=[-1 for i in range(n)]        res=[]        dfs(0,[])        return res

非递归回溯
http://www.cnblogs.com/zuoyuan/p/3747658.html

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