https://tigerisland.blog.csdn.net/article/details/54755895
C++ 代码
方法一:
这是一个将二维矩阵元素映射为一维序列的问题。这个问题的关键是,
输出一个元素(假设位于<row,col>)后,下一个输出元素的位置如何计算?
只要找出一套计算下一个输出元素下标的规则,问题就解决了。
方法二:
如果找到下标从左上角到右下角的变化规律,问题就解决了。
1.
/* 201412-2 Z字形扫描 */
#include <iostream>
using namespace std;
const int EAST = 0;
const int SOUTH = 1;
const int SOUTHWEST = 2;
const int NORTHEAST = 3;
struct {
int drow;
int dcol;
} direct[] = {{0, 1}, {1, 0}, {1, -1}, {-1, 1}};
const int N = 500;
int a[N][N];
int main()
{
int n;
// 输入数据
cin >> n;
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
cin >> a[i][j];
// Z字形扫描,输出结果
int row=0, col=0, next=EAST;
cout << a[row][col];
while(row != n - 1 || col != n - 1) {
row += direct[next].drow;
col += direct[next].dcol;
cout << " " << a[row][col];
if(next == EAST && row == 0)
next = SOUTHWEST;
else if(next == EAST && row == n - 1)
next = NORTHEAST;
else if(next == SOUTH && col == 0)
next = NORTHEAST;
else if(next == SOUTH && col == n - 1)
next = SOUTHWEST;
else if(next == SOUTHWEST && row == n - 1)
next = EAST;
else if(next == SOUTHWEST && col == 0)
next = SOUTH;
else if(next == NORTHEAST && col == n - 1)
next = SOUTH;
else if(next == NORTHEAST && row == 0)
next = EAST;
}
cout << endl;
return 0;}
2.
#include <iostream>
using namespace std;
const int N = 500;
int a[N][N];
int main()
{
int n, x, y;
// 输入数据
cin >> n;
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
cin >> a[i][j];
// 输出左上三角
x = 0;
y = 0;
for(int i=0; i<n; i++)
if(i & 1) {
for(int j=0; j<i; j++)
cout << a[x++][y--] << " ";
cout << a[x++][y] << " ";
} else {
for(int j=0; j<i; j++)
cout << a[x--][y++] << " ";
cout<< a[x][y++] << " ";
}
// 输出右下三角
if(n & 1)
y--, x++;
else
y++, x--;
for(int i=n-2; i>0; i--)
if(i & 1) {
for(int j=0; j<i; j++)
cout << a[x++][y--] << " ";
cout << a[x][y++] << " ";
} else {
for(int j=0; j<i; j++)
cout << a[x--][y++] << " ";
cout << a[x++][y] << " ";
}
if(n!=1)
cout << a[n-1][n-1] << endl;
return 0;
}