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法律研究以螺旋方式打印矩阵\n
此算法用于以螺旋方式打印数组元素。首先从第一行开始，打印整个内容，然后打印最后一列，然后是最后一行，依此类推，因此以螺旋方式打印元素。
此算法的时间复杂度为 O(MN)，其中 M 是行数，N 是列数。
输入输出
Input: The matrix: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Output: Contents of an array as the spiral form 1 2 3 4 5 6 12 18 17 16 15 14 13 7 8 9 10 11 15 16
算法
dispSpiral(mat, m, n)
输入：矩阵 mat，行和列 m 和 n。
输出：以螺旋方式打印矩阵的元素。
Begin currRow := 0 and currCol := 0 while currRow and currCol are in the matrix range, do for i in range currCol and n-1, do display mat[currRow, i] done increase currRow by 1 for i in range currRow and m-1, do display mat[i, n-1] done decrease n by 1 if currRow < m, then for i := n-1 down to currCol, do display mat[m-1, i] done decrease m by 1 if currCol < n, then for i := m-1 down to currRow, do display mat[i, currCol] done increase currCol by 1 done End
示例
#include <iostream>
#define ROW 3
#define COL 6
using namespace std;
int array[ROW][COL] = {
{1, 2, 3, 4, 5, 6},
{7, 8, 9, 10, 11, 12},
{13, 14, 15, 16, 17, 18}
};
void dispSpiral(int m, int n) {
int i, currRow = 0, currCol = 0;
while (currRow < ROW && currCol <COL) {
for (i = currCol; i < n; i++) { //print the first row normally
cout << array[currRow][i]<<" ";
}
currRow++; //point to next row
for (i = currRow; i < m; ++i) { //Print the last column
cout << array[i][n-1]<<" ";
}
n--; //set the n-1th column is current last column
if ( currRow< m) { //when currRow is in the range, print the last row
for (i = n-1; i >= currCol; --i) {
cout << array[m-1][i]<<" ";
}
m--; //decrease the row range
}
if (currCol <n) { //when currCol is in the range, print the fist column
for (i = m-1; i >= currRow; --i) {
cout << array[i][currCol]<<" ";
}
currCol++;
}
}
}
int main() {
dispSpiral(ROW, COL);
}输出
1 2 3 4 5 6 12 18 17 16 15 14 13 7 8 9 10 11 15 16
更新于： 17-Jun-2020
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