leetcode No51. N-Queens

Question:

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:

Algorithm:

由鹊巢原理，每一层一定有一个Queen，所以不妨从第一行往下试探，用一个数组cur存放Queen的列标。
首先需要一个判断是否合法的函数isVaild判断是否和之前的Queen冲突，由于是逐行试探，所以不需要判断行号，只需判断列号和斜线。
从第一行往下试探，如果是合法的，继续往下试探，一直到n都是合法的那么就是一个解；试探过程中一旦不合法，停止。

Accepted Code：

class Solution {
vector<vector<string>> res;
public:
vector<vector<string>> solveNQueens(int n) {    //有很多解
vector<int> v(n);
result(v,n,0);
return res;
}
private:
bool isValid (vector<int> &cur,int c)    //cur为记录的列数数组，c为当前行数,cur[c]为当前列数
{
for(int i=0;i<c;i++)   //只需判断列数和斜线
{
if(cur[i]==cur[c]||abs(cur[c]-cur[i])==abs(c-i))
return false;
}
return true;
}

void result(vector<int> &cur,int n,int c)
{
if(c==n)      //因为c是从0开始的，所以c==n时，已经有n个数了
{
vector<string> temp(n,string(n,'.'));
for(int i=0;i<n;i++)
{
temp[i][cur[i]]='Q';
}
res.push_back(temp);
}
for(cur[c]=0;cur[c]<n;cur[c]++)
{
if(isValid(cur,c))
{
cur.push_back(cur[c]);
result(cur,n,c+1);
}
}
}
};