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Copy pathcirclePrintMatrix.cpp
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99 lines (95 loc) · 2.07 KB
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/*转圈打印矩阵，例如矩阵
1 2 打印结果为 1 2 4 3
3 4
分析：如果按照下标偏移来控制打印的话太过麻烦，
所以维护2个顶点坐标，每次外围打印一圈数字。
要求空间复杂度为O(1)
*/
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// 打印边缘一圈的函数
void printpEdge(vector<vector<int>> m, int row1, int clo1, int row2, int clo2)
{
if (row1 > row2 || clo1 > clo2){
cerr << " row or clo wrong" << endl;
return;
}
if (clo1 == clo2){
// 列标相等 说明只有1列
for (int i = row1; i <= row2; i++){
cout << m[i][clo1] << " ";
}
}
else if (row1 == row2){
// 行标相等 说明只有1行
for (int i = clo1; i <= clo2; i++){
cout << m[row1][i] << " ";
}
}
else{
// 注意顶角4元素不要重复打印
// 从左到右打印第row1行
for (int i = clo1; i <= clo2; i++){
cout << m[row1][i] << " ";
}
// 从上到下打印第clo2列
for (int i = row1 + 1; i <= row2; i++){
cout << m[i][clo2] << " ";
}
// 从右到左打印第row2行
for (int i = clo2-1; i >= clo1; i--){
cout << m[row2][i] << " ";
}
// 从下到上打印第clo1列
for (int i = row2 - 1; i >= row1 + 1; i--){
cout << m[i][clo1] << " ";
}
}
}
// 转圈打印矩阵
void circleMatrix(vector<vector<int>> m)
{
if (m.empty()){
cerr << "martix is null";
return;
}
int row1 = 0;
int clo1 = 0;
int row2 = m.size() - 1;
int clo2 = m[0].size() - 1;
while (row1 <= row2 && clo1 <= clo2){
printpEdge(m, row1, clo1, row2, clo2);
row1++;
clo1++;
row2--;
clo2--;
}
}
int main(){
vector<int> tmp;
vector<vector<int>> m;
for (int i = 1; i < 15; i++)
tmp.push_back(i);
for (int i = 0; i < 4; i++)
{
if (i%2==0)
m.push_back(tmp);
else{
reverse(tmp.begin(),tmp.end());
m.push_back(tmp);
}
}
cout << "matrix content is:" << endl;
for (int i = 0; i < m.size(); i++){
for (int j = 0; j < m[0].size(); j++){
cout << m[i][j] << " ";
}
cout << endl;
}
cout << endl << "circle print result:" << endl;
circleMatrix(m);
system("pause");
return 0;
}