面试题29. 顺时针打印矩阵

//分圈方式打印矩阵，一个点代码左上，一个点代码右下，把第一圈剥离出来，然后打印，再去打印子矩阵
class Solution {
    static int i = 0;
    int[] res = new int[0];
    public int[] spiralOrder(int[][] matrix) {
        if(matrix == null || matrix.length == 0 ||matrix[0].length == 0) return res;
        int m = matrix.length;
        int n = matrix[0].length;
        res = new int [m*n];
        //左上角点坐标
        int lR = 0;
        int lC = 0;
        int tR = m-1;
        int tC = n-1;
        i = 0;
        while(lR <= tR && lC <= tC){
            printMatrix(matrix,lR,lC,tR,tC);
            lR++;
            lC++;
            tR--;
            tC--;
        }
        return res;
    }
    //打印圈
    void printMatrix(int[][] matrix,int lR,int lC,int tR,int tC){
        //如果只有一行 lR == tR
        if(lR == tR){
            while(lC <= tC){
                res[i] = matrix[lR][lC];
                i++;
                lC++;
            } 
            return;
        }
        //如果只有一列 lC == tC
        else if(lC == tC){
            while(lR <= tR){
                res[i++] = matrix[lR][lC];
                lR++;
            } 
            return;
        }
        else{
            //一圈，循环打印
            int curR = lR;
            int CurC = lC;
            //打印行,列坐标++
            while(CurC != tC){
                res[i] = matrix[lR][CurC];
                i++;
                CurC++;
            }
            //打印列,行坐标++
            while(curR != tR){
                res[i] = matrix[curR][tC];
                i++;
                curR++;
            }
            //打印行，列坐标--
            while(CurC != lC){
                res[i] = matrix[tR][CurC];
                i++;
                CurC--;
            }
            //打印列，行坐标--
            while(curR != lR){
                res[i] = matrix[curR][lC];
                i++;
                curR--;
            }
        }
    }
}