leetcode上面难度hard题_leetcode上面的题,hard难度怎么样?

bool isvalid(int row,int col,int N,vector<int> &record)//判断是否能够放置
{
    for(int i=0;i<N;i++)
    {
        if(record[i] == col || abs(i-row) == abs(record[i]-col))
            return false;
    }
    return true;
}

void print(vector<vector<string>> &result,vector<int> &record,int N)//输出
{
    ostringstream re;
    vector<string> temp;
    for(int i=0;i<N;i++)
    {
        for(int j=0;j<N;j++)
        {
            if(record[i] == j)
                re<<'Q';
            else
                re<<'.';
        }
        temp.push_back(re.str());
        re.str("");
    }
    result.push_back(temp);
}

• 非递归的主体函数实现

  vector<vector<string>> solveNQueens(int n) {
      vector<vector<string>> result;
      int row=0,col=0;
      vector<int> record(n,-10000);
      while(row<n)
      {
          while(col<n)
          {
              if(isvalid(row,col,n,record))
              {
                  record[row] = col;
                  col = 0;
                  break;
              }
              else
              {
                  col++;
              }
          }
          if(record[row] == -10000)//若这一行没有放置
          {
              if(row==0)//退到第一行，程序结束
                  break;
              else
              {
                  --row;   //回退到上一行
                  col=record[row] +1;//上一行往右移一列
                  record[row] = -10000;//清除上一行的皇后
                  continue;
              }
          }
          if(row == n-1)
          {
              print(result,record,n);
              col = record[row]+1;//往右移一列
              record[row] = -10000;//清除本行的皇后
              continue;
          }
          ++row;
      }
      return result;
  }