剑指offer系列13：顺时针打印矩阵
这种题，必然要一圈一圈循环，因此，考虑两个问题：1.如何终止循环 2.如何打印一圈
我今天偷懒了， main函数是我从答案搬来的。
1 #include<iostream> 2 #include<vector> 3 using namespace std; 4 class Solution { 5 public: 6 vector<int> printMatrix(vector<vector<int> > matrix) { 7 int row = matrix.size(); 8 int col = matrix[0].size(); 9 int left = 0, right = col-1;//列标 10 int top = 0, bot = row-1;//行标 11 vector<int> result; 12 while (left <=right&&top <= bot) 13 { 14 //打印上面一行，列是变量 15 for (int i = left; i <= right; i++) 16 { 17 result.push_back(matrix[top][i]); 18 } 19 //打印右边一列,行是变量 20 if (bot > top) 21 { 22 for (int i = top+1; i <=bot; i++) 23 { 24 result.push_back(matrix[i][right]); 25 } 26 } 27 //打印下边一行，列是变量 28 if (bot > top&&right > left) 29 { 30 for (int i = right-1; i >=left; i--) 31 { 32 result.push_back(matrix[bot][i]); 33 34 } 35 } 36 //打印左边一列，行是变量 37 if (bot-top>1 && right > left) 38 { 39 for (int i = bot-1; i >top; i--) 40 { 41 result.push_back(matrix[i][left]); 42 } 43 } 44 top++; 45 right--; 46 bot--; 47 left++; 48 49 } 50 return result; 51 } 52 }; 53 int main() 54 { 55 int a1[] = { 1, 2, 3, 4 }; 56 int a2[] = { 5, 6, 7, 8 }; 57 int a3[] = { 9, 10, 11, 12 }; 58 int a4[] = { 13, 14, 15, 16 }; 59 // int a1[] = { 1 }; 60 // int a2[] = { 2 }; 61 // int a3[] = { 3 }; 62 // int a4[] = { 4 }; 63 64 vector<int> vec1(a1, a1 + 4); 65 vector<int> vec2(a2, a2 + 4); 66 vector<int> vec3(a3, a3 + 4); 67 vector<int> vec4(a4, a4 + 4); 68 69 70 vector< vector<int> > vec; 71 vec.push_back(vec1); 72 vec.push_back(vec2); 73 vec.push_back(vec3); 74 vec.push_back(vec4); 75 76 Solution solu; 77 vector<int> res = solu.printMatrix(vec); 78 for (int i = 0; i < res.size(); i++) 79 { 80 cout << res[i] << " "; 81 } 82 return 0; 83 }
