难度系数 ⭐⭐
时间限制 C/C++ 1秒,其他语言2秒 空间限制:C/C++ 32M,其他语言64M
题目内容 输入一个矩阵,按照从外向里以顺时针的顺序依次打印出每一个数字,例如:
如果输入如下4 X 4矩阵: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
则依次打印出数字:1,2,3,4,8,12,16,15,14,13,9,5,6,7,11,10.
思路 使用递归。如果当前矩阵行、列数都至少是2,则可以打印一圈,打印完成后,行、列数各减少2,递归计算剩余部分。
package nowcoder;
import java.util.ArrayList;
public class No13 {
public static ArrayList<Integer> printMatrix(int[][] matrix) {
ArrayList<Integer> res = new ArrayList<>();
int left = 0;
int right = matrix[0].length;
int top = 0;
int bottom = matrix.length;
printRes(matrix, left, right, top, bottom, res);
return res;
}
public static void printRes(int[][] matrix, int left, int right, int top, int bottom, ArrayList<Integer> res){
if (left < right - 1 && top < bottom - 1){
for (int l = left; l < right - 1; l++) res.add(matrix[top][l]);
for (int t = top; t < bottom - 1; t++) res.add(matrix[t][right-1]);
for (int r = right - 1; r > left; r--)res.add(matrix[bottom-1][r]);
for (int b = bottom - 1; b > top; b--) res.add(matrix[b][left]);
printRes(matrix, left + 1, right - 1, top + 1, bottom - 1, res);
} else if (left < right - 1 && top == bottom - 1){
for (int l = left; l < right; l++) res.add(matrix[top][l]);
} else if (left == right - 1 && top < bottom - 1) {
for (int t = top; t < bottom; t++) res.add(matrix[t][left]);
}
else if (left == right - 1 && top == bottom - 1)
res.add(matrix[top][left]);
}
public static void main(String[] args){
int[][] matrix = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16}};
// int[][] matrix = {{1}};
System.out.println(printMatrix(matrix));
}
}