在 R 行 C 列的矩阵上，我们从 (r0, c0) 面朝东面开始
这里，网格的西北角位于第一行第一列，网格的东南角位于最后一行最后一列。
现在，我们以顺时针按螺旋状行走，访问此网格中的每个位置。
每当我们移动到网格的边界之外时，我们会继续在网格之外行走（但稍后可能会返回到网格边界）。
最终，我们到过网格的所有 R * C 个空间。
按照访问顺序返回表示网格位置的坐标列表。
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用程序求阿基米德线的一部分
暴力解法参考：
螺旋矩阵 III（failed）
本方法：
如果矩阵坐标按照规则增加越界，自动寻找下一个矩阵坐标范围内的起始点
示例 1：
输入：R = 1, C = 4, r0 = 0, c0 = 0
输出：[[0,0],[0,1],[0,2],[0,3]]
示例 2：
输入：R = 5, C = 6, r0 = 1, c0 = 4
输出：[[1,4],[1,5],[2,5],[2,4],[2,3],[1,3],[0,3],[0,4],[0,5],[3,5],[3,4],[3,3],[3,2],[2,2],[1,2],[0,2],[4,5],[4,4],[4,3],[4,2],[4,1],[3,1],[2,1],[1,1],[0,1],[4,0],[3,0],[2,0],[1,0],[0,0]]
提示：
1 <= R <= 1001 <= C <= 1000 <= r0 < R0 <= c0 < C
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <deque>
#include <map>
using namespace std;
class Solution
{
private:
int limit_up, limit_down, limit_left, limit_right;
int rmax, cmax;
bool confine = false;
int count = 1;
public:
vector<vector<int>> spiralMatrixIII(int R, int C, int r0, int c0)
{
rmax = R - 1;
cmax = C - 1;
limit_up = r0, limit_down = r0, limit_left = c0, limit_right = c0;
vector<pair<int, int>>temp;
vector<vector<int>> ans;
temp.push_back({ r0,c0 });
cout << count << ": " << r0 << " " << c0 << endl;
int s;
for (int i = 1; i < 5; i++)
{
if (temp.size() == R * C)
break;
s = i;
switch (s)
{
case 1:
if (confine)
{
if (limit_right != cmax)
{
c0 = limit_right + 1;
limit_right++;
confine = 0;
count++, cout << count << ": " << r0 << " " << c0 << endl, temp.push_back({ r0,c0 });
}
else
{
c0 = limit_right;
}
break;
}
else
while (c0 < C )
{
c0++;
//step1:
if (c0 < C)
count++, cout << count << ": " << r0 << " " << c0 << endl, temp.push_back({ r0,c0 });
if (c0 - limit_right == 1 && limit_right < C-1)
{
limit_right = c0;
break;
}
else
{
trans(r0, c0, i);
if (i != s)
break;
}
}
break;
case 2:
if (confine)
{
if (limit_down!=rmax)
{
r0 = limit_down + 1;
limit_down++;
confine = 0;
count++,cout << count << ": " << r0 << " " << c0 << endl, temp.push_back({ r0,c0 });
}
else
{
r0 = limit_down;
}
break;
}
else
while (r0 < R )
{
r0++;
//step2:
if (r0 < R)
count++,cout << count << ": " << r0 << " " << c0 << endl, temp.push_back({ r0,c0 });
if (r0 - limit_down == 1 && limit_down < R-1)
{
limit_down = r0;
break;
}
else
{
trans(r0, c0, i);
if (i != s)
break;
}
}
break;
case 3:
if (confine)
{
if (limit_left!=0)
{
c0 = limit_left - 1;
limit_left--;
confine = 0;
count++, cout << count << ": " << r0 << " " << c0 << endl, temp.push_back({ r0,c0 });
}
else
{
c0 = 0;
}
break;
}
else
while (c0 > -1)
{
c0--;
//step3:
if (c0 > -1)
count++,cout << count << ": " << r0 << " " << c0 << endl, temp.push_back({ r0,c0 });
if (limit_left - c0 == 1 && limit_left > 0)
{
limit_left = c0;
break;
}
else
{
trans(r0, c0, i);
if (i != s)
break;
}
}
break;
case 4:
if (confine)
{
if (limit_up!=0)
{
r0 = limit_up - 1;
limit_up--;
confine = 0;
count++, cout << count << ": " << r0 << " " << c0 << endl, temp.push_back({ r0,c0 });
}
else
{
r0 = 0;
}
break;
}
else
while (r0 > -1)
{
r0--;
if (r0 > -1)
count++,cout << count << ": " << r0 << " " << c0 << endl, temp.push_back({ r0,c0 });
if (limit_up - r0 == 1 && limit_up > 0)
{
limit_up = r0;
break;
}
else
{
trans(r0, c0, i);
if (i != s)
break;
}
}
break;
}
if (i != s)
i--;
if (i == 4)
i = 0;
}
for (auto i = temp.begin(); i !=temp.end(); i++)
{
pair<int, int>te = *i;
ans.push_back({ te.first,te.second });
}
return ans;
}
void trans( int &r0, int &c0,int &i)
{
if (c0>cmax && i == 1)
{
c0 = cmax;
confine = true;
i++;
}
else if (r0>rmax && i == 2)
{
r0 = rmax;
confine = true;
i++;
}
else if (c0<0 && i == 3)
{
c0 = 0;
confine = true;
i++;
}
else if (r0<0 && i == 4)
{
r0 = 0;
confine = true;
i++;
}
}
};
int main()
{
int R = 1, C = 4, r0 = 0, c0 = 0;
Solution s;
s.spiralMatrixIII(R, C, r0, c0);
return 0;
}