[LeetCode] Spiral Matrix

这个题目的意思比较简单,就是螺旋式的输出数组的数据,[
[

 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]
You should return [1,2,3,6,9,8,7,4,5].

我自己的解法比较直接,就是从最外层向内层逐步递归这种方式不需要额外的存储空间,但是边界控制比较繁琐,代码如下:

    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> rs = new LinkedList<Integer>();
        int m, n;//row and column
        if(matrix == null || matrix.length == 0)
            return rs;
        else {
            m = matrix.length;
            n = matrix[0].length;
            getSpiralData(matrix, 0, 0, m, n, rs);
        }
        return rs;
    }

    /**
     * 循环获取数据
     * @param matrix data matrix
     * @param startX 起始位置
     * @param startY
     * @param maxM max row
     * @param maxN max column
     * @param rs result list
     */
    public void getSpiralData(int[][] matrix, int startX, int startY, int maxM, int maxN, List<Integer> rs){
        if(maxM <= 0 || maxN <= 0){
            return;
        } else {
            int x= startX, y = startY;
            if(maxM ==1){
                while (y < startY + maxN){
                    rs.add(matrix[x][y]);
                    y++;
                }
                return;
            }
            if(maxN ==1){
                while (x < startX + maxM){
                    rs.add(matrix[x][y]);
                    x++;
                }
                return;
            }

            while (y <= startY + maxN-1){
                rs.add(matrix[x][y]);
                y++;
            }
            x = x + 1;
            y = startY + maxN -1;
            while (x <= startX + maxM-1){
                rs.add(matrix[x][y]);
                x++;
            }
            y = y-1;
            x = startX + maxM -1;
            while (y >= startY){
                rs.add(matrix[x][y]);
                y--;
            }
            x = x-1;
            y = startY;
            while (x > startX){
                rs.add(matrix[x][y]);
                x--;
            }
            getSpiralData(matrix,startX+1, startY+1, maxM-2, maxN-2, rs);
        }
    }

http://www.cnblogs.com/remlostime/archive/2012/11/18/2775708.html

class Solution {
private:
    int step[4][2];
    vector<int> ret;
    bool canUse[100][100];
public:
    void dfs(vector<vector<int> > &matrix, int direct, int x, int y)
    {
        for(int i = 0; i < 4; i++)
        {
            int j = (direct + i) % 4;
            int tx = x + step[j][0];
            int ty = y + step[j][1];
            if (0 <= tx && tx < matrix.size() && 0 <= ty && ty < matrix[0].size() && canUse[tx][ty])
            {
                canUse[tx][ty] = false;
                ret.push_back(matrix[tx][ty]);                
                dfs(matrix, j, tx, ty);               
            }            
        }
    }
    
    vector<int> spiralOrder(vector<vector<int> > &matrix) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        step[0][0] = 0;
        step[0][1] = 1;
        step[1][0] = 1;
        step[1][1] = 0;
        step[2][0] = 0;
        step[2][1] = -1;
        step[3][0] = -1;
        step[3][1] = 0;
        ret.clear();
        memset(canUse, true, sizeof(canUse));
        dfs(matrix, 0, 0, -1);
        
        return ret;
    }
};

[LeetCode] Spiral Matrix

原文：http://blog.csdn.net/youmengjiuzhuiba/article/details/44617793