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Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
For example,
Given the following matrix:
[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]
You should return [1,2,3,6,9,8,7,4,5].
用left, top, right, bottom记录边界条件。
1 vector<int> spiralOrder(vector<vector<int> > &matrix) 2 { 3 vector<int> ret; 4 int m = matrix.size(); 5 if (m <= 0) 6 return ret; 7 int n = matrix[0].size(), i = 0, j = 0; 8 int left = 0, right = n - 1, top = 0, bottom = m - 1; 9 10 while (true) 11 { 12 i = top, j = left; 13 while (j <= right) 14 ret.push_back(matrix[i][j++]); 15 if (++top > bottom) break; 16 17 i = top, j = right; 18 while(i <= bottom) 19 ret.push_back(matrix[i++][j]); 20 if (--right < left) break; 21 22 i = bottom, j = right; 23 while (j >= left) 24 ret.push_back(matrix[i][j--]); 25 if (--bottom < top) break; 26 27 i = bottom, j = left; 28 while (i >= top) 29 ret.push_back(matrix[i--][j]); 30 if (++left > right) break; 31 } 32 33 return ret; 34 }
Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.
For example,
Given n = 3,
You should return the following matrix:
[ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ] ]
1 vector<vector<int> > generateMatrix(int n) 2 { 3 vector<vector<int> > matrix; 4 if (n <= 0) 5 return matrix; 6 int left = 0, right = n - 1, top = 0, bottom = n - 1, i, j, k = 1; 7 vector<int> line(n); 8 for (i = 0; i < n; i++) 9 matrix.push_back(line); 10 while (true) 11 { 12 i = top, j = left; 13 while (j <= right) 14 matrix[i][j++] = k++; 15 if (++top > bottom) break; 16 17 i = top, j = right; 18 while(i <= bottom) 19 matrix[i++][j] = k++; 20 if (--right < left) break; 21 22 i = bottom, j = right; 23 while (j >= left) 24 matrix[i][j--] = k++; 25 if (--bottom < top) break; 26 27 i = bottom, j = left; 28 while (i >= top) 29 matrix[i--][j] = k++; 30 if (++left > right) break; 31 } 32 33 return matrix; 34 }
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原文地址：http://www.cnblogs.com/ym65536/p/4177092.html
