885. Spiral Matrix III（python+cpp）

题目：

On a 2 dimensional grid with R rows and C columns, we start at (r0, c0) facing east.
Here, the north-west corner of the grid is at the first row and column, and the south-east corner of the grid is at the last row and column.
Now, we walk in a clockwise spiral shape to visit every position in this grid.
Whenever we would move outside the boundary of the grid, we continue our walk outside the grid (but may return to the grid boundary later.)
Eventually, we reach all R * C spaces of the grid.
Return a list of coordinates representing the positions of the grid in the order they were visited.
Example 1:
Input: R = 1, C = 4, r0 = 0, c0 = 0 
Output: [[0,0],[0,1],[0,2],[0,3]]
Example 2:
Input: R = 5, C = 6, r0 = 1, c0 = 4 
Output:
[[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]]
Note:
1 <= R <= 100
1 <= C <= 100
0 <= r0 < R
0 <= c0 < C

解释：
记录每次该走的步数以及要走的步数以及要走的方向。
步数是1,1,2,2,3,3,4,4……这样的形式，行走方向是正正负负正正负负这样的形式，方向在 i 和j走完之后都会反向，所以可以用一个sign来表示行走方向的正负，走完一行一列以后，sign反向，每一个sign中，所需要走的列数（i）和行数（j）是相等的，为stepSize，走完一行一列以后，stepSize需要+1。
python代码：

class Solution:
    def spiralMatrixIII(self, R, C, r0, c0):
        """
        :type R: int
        :type C: int
        :type r0: int
        :type c0: int
        :rtype: List[List[int]]
        """
        i,j=r0,c0
        sign=1
        result=[[r0,c0]]
        stepSize=1
        while len(result)<R*C:
            for _ in range(stepSize):
                j+=sign
                if i>=0 and i<R and j>=0 and j<C:
                    result.append([i,j])
            for _ in range(stepSize):
                i+=sign
                if i>=0 and i<R and j>=0 and j<C:
                    result.append([i,j])
            sign*=-1
            stepSize+=1
        return result

c++代码：

class Solution {
public:
    vector<vector<int>> spiralMatrixIII(int R, int C, int r0, int c0) {
        int i=r0,j=c0;
        int stepSize=1;
        int sign=1;
        vector<vector<int>>result={{r0,c0}};
        while(result.size()<R*C)
        {
            for(int k=0;k<stepSize;k++)
            {
                j+=sign;
                if(i>=0 &&i<R &&j>=0 && j<C)
                    result.push_back({i,j});
            }
            for(int k=0;k<stepSize;k++)
            {
                i+=sign;
                if(i>=0 &&i<R &&j>=0 && j<C)
                    result.push_back({i,j});
            }
            sign*=-1;
            stepSize++;
        }
        return result;
    }
};

*总结：

885. Spiral Matrix III（python+cpp）

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