N-Queens

Description:

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.."]
]

Code:

class Solution {
public:
int x[100];
bool place(int k)//考察皇后k放置在x[k]列是否发生冲突
{
    int i;
    for(i=1;i<k;i++)
        if(x[k]==x[i]||abs(k-i)==abs(x[k]-x[i]))
            return false;
        return true;
}

vector<string> toVectorString(int n)
{
    vector<string>result;
    for (int i = 1; i <= n; ++i)
    {
        string str(n,'.');
        str[x[i]-1]='Q';
        result.push_back(str);
    }
    return result;
}

    vector<vector<string>> solveNQueens(int n) {
    vector< vector<string> >result;
   for(int i=1;i<=n;i++)
        x[i]=0;
    int k=1;
    while(k>=1)
    {//x[1]...x[n]表示第k个皇后在第x[k]列
        x[k]=x[k]+1;   //在下一列放置第k个皇后
        while(x[k]<=n&&!place(k))
            x[k]=x[k]+1;//搜索下一列
        if(x[k]<=n&&k==n)//得到一个输出
        {
            vector<string>temp = toVectorString(n);
            result.push_back(temp);
        }
        else if(x[k]<=n&&k<n)
            k=k+1;//放置下一个皇后
        else
        {
            x[k]=0;//重置x[k],回溯
            k=k-1;
        }
    }
    return result;
    }
};