leetcode -- N-Queens I&II，经典回溯，再看

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