题目的要点是:怎么快速判断这次的皇后和以前的是不冲突的。
class Solution {
public:
vector<int> hor;
vector<int> ver;
vector<int> rightline;
vector<int> leftline;
int len;
vector<vector<string>> result;
vector<vector<string> > solveNQueens(int n) {
if(n<=0)
return result;
vector<string> temp;
Init(n,temp);
SolveQueens(temp,n);
return result;
}
void Init(int n,vector<string>&temp)
{
len =n;
hor.resize(n);
ver.resize(n);
rightline.resize(2*n);
leftline.resize(2*n);
string str(n,‘.‘);
for(int i=0; i<n;i++)
temp.push_back(str);
}
void SolveQueens(vector<string>& temp,int num)
{
if(num==0)
{
result.push_back(temp);
return ;
}
int i=0;
i = len-num;
for(int j=0; j<len;j++)
{
if(CheckValid(i,j))
{
temp[i][j] = ‘Q‘;
SetFlag(i,j);
SolveQueens(temp,num-1);
ClearFlag(i,j);
temp[i][j] = ‘.‘;
}
}
}
void SetFlag(int i,int j)
{
hor[i] = 1;
ver[j] = 1;
int index = GetIndex(i,j) ;
rightline[index] = 1;
leftline[i+j] = 1;
}
void ClearFlag(int i,int j)
{
hor[i] = 0;
ver[j] = 0;
int index = GetIndex(i,j);
rightline[index] = 0;
leftline[i+j] = 0;
}
int GetIndex(int i,int j)
{
int index = 0;
if(i>=j)
{
index = i-j;
}
else
{
index = j-i+len;
}
return index;
}
bool CheckValid(int i,int j)
{
if(hor[i] == 1|| ver[j] == 1)
return false;
int index = GetIndex(i,j);
if(rightline[index] == 1|| leftline[i+j] == 1)
return false;
return true;
}
};
原文:http://www.cnblogs.com/xgcode/p/4230389.html