N-Queens leetcode
The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

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 indicate a queen and an empty space respectively.
For example,
There exist two distinct solutions to the 4-queens puzzle:
[
[".Q..", // Solution 1
"...Q",
"Q...",
"..Q."], ["..Q.", // Solution 2
"Q...",
"...Q",
".Q.."]
]
解题思路:
使用回溯法解决N皇后问题,是常见的解决方法。分N步放置皇后,由于皇后之间不能同行、同列、同斜线,所以每次放置皇后的时候,都要考虑是否与已有的皇后“冲突”,如果冲突,则改变位置,直至所有皇后放置完毕。
测试皇后冲突的函数:isValid();
本文思路比较取巧,使用a[n]记录N皇后位置。根据皇后放置规则可知,每一行有且只有一个皇后,所以从第一行到第N行,依次放置第n个皇后。a[n]记录第n个皇后所在列数。即第i个皇后放置在第i行第a[i]列。
参考:http://blog.csdn.net/feixiaoxing/article/details/6877965
具体实现如下(AC 36ms):
class Solution {
public:
vector<vector<string> > re;
//测试在第row行,第row列放置皇后是否有效
int isValid(int *a, int n, int row, int col)
{
int tmpcol=0;
for(int tmprow=0;tmprow<row;tmprow++)
{
tmpcol = a[tmprow];
if(tmpcol == col)// 同列
return 0;
if((tmpcol-col) == (tmprow - row))// 在同一右斜线
return 0;
if((tmpcol-col) == (row - tmprow))// 在同一左斜线
return 0;
}
return 1;
}
void PrintN(int *a, int n)
{
vector<string> tmps;
for(int i=0;i<n;i++)
{
string s(n,'.');
s[a[i]]='Q';
tmps.push_back(s);
}
re.push_back(tmps);
}
void n_queens(int *a,int n, int index)
{
for(int i=0;i<n;i++)
{
if(isValid(a,n,index,i))
{
a[index]=i;
if(index == n-1)
{
PrintN(a,n);
a[index]=0;
return;
}
n_queens(a,n,index+1);
a[index]=0;
}
}
}
vector<vector<string> > solveNQueens(int n) {
int *a = new int[n];
memset(a,0,sizeof(int)*n);
n_queens(a,n,0);
return re;
}
};
N皇后个数对应解的个数:(验证程序)
n solution(n)
1 1
2 0
3 0
4 2
5 10
6 4
7 40
8 92
9 352
10 724
11 2680
12 14200
13 73712
14 365596
15 2279184
16 14772512
17 95815104
18 666090624
19 4968057848
20 39029188884
21 314666222712
22 2691008701644
23 24233937684440
24 227514171973736
25 2207893435808352
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