面试题29

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顺时针打印矩阵

注意：
(1)边界条件
(2)矩阵只有1行、1列、1行1列这三种情况都要考虑

#include<stdio.h>int A[4][3] = { { 1,2,3 },{ 4,5,6 },{ 7,8,9 } ,{ 10,11,12 } };int row = sizeof(A) / sizeof(A[0]);int column = sizeof(A[0]) / sizeof(A[0][0]);void CirclePrintMatric(int row, int column, int start) {    int endrow = row - start - 1;    int endcolumn = column - start - 1;    for (int i = start; i <= endcolumn; i++) {    //从左往右打印        printf("%d ", A[start][i]);    }    if (start < endrow) {             //从上往下打印        for (int i = start+1; i <= endrow; i++) {            printf("%d ", A[i][endcolumn]);        }    }    if (start < endcolumn&&start < endrow) {     //从右往左打印        for (int i = endcolumn - 1; i >= start; i--) {            printf("%d ", A[endrow][i]);        }    }    if (start < endcolumn&&start < endrow - 1) {       //从下往上打印        for (int i = endrow - 1; i > start; i--) {            printf("%d ", A[i][start]);        }    }}void PrintMatric(int row,int column) {    if (A == NULL||row<=0||column<=0)        return;    int start = 0;    while (row > 2 * start&&column > 2 * start) {        CirclePrintMatric(row, column, start);        start++;    }}int main() {    PrintMatric(row,column);    return 0;}

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