leetcode 54|59. Spiral Matrix 1|2
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54. Spiral Matrix
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].
旋转读取二维矩阵,x,y是每一圈的左上角(定点坐标)的横纵坐标,row是每一圈的行数,col是每一圈的列数。
每一圈扫完,定点坐标往右下移动,row-2, col-2
class Solution {public: void help(vector<vector<int>>& matrix, vector<int> &ret, int x, int y, int row, int col) { if (row == 0 || col == 0) return; else if (col == 1) { for (int i = x; i < x + row; i++) ret.push_back(matrix[i][y]); return; } else if (row == 1) { for (int i = y; i < y + col; i++) ret.push_back(matrix[x][i]); return; } else { for (int i = y; i < y + col; i++) ret.push_back(matrix[x][i]); for (int i = x + 1; i < x + row - 1; i++) ret.push_back(matrix[i][y + col - 1]); for (int i = y + col - 1; i >= y; i--) ret.push_back(matrix[x + row - 1][i]); for (int i = x + row - 2; i > x; i--) ret.push_back(matrix[i][y]); help(matrix, ret, x + 1, y + 1, row - 2, col - 2); } return; } vector<int> spiralOrder(vector<vector<int>>& matrix) { vector<int> ret; if (matrix.size() == 0) return ret; int row = matrix.size(); int col = matrix[0].size(); help(matrix, ret, 0, 0, row, col); return ret; }};
59. Spiral Matrix II
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 ]
]
way-1:找规律输出就行
class Solution {public: void Matrix(int n, int pos, int count, vector<vector<int>> &result) { //上面一行输入 for (int i = pos; i < n - pos; i++) result[pos][i] = count ++; //右边一列 for(int i = pos + 1; i < n - pos - 1; i++) result[i][n-1-pos] = count ++; if (count > n * n) //上面和右边填入之后需要判断,因为可能只有一个坑,而后面可能会重新覆盖 return; //下面一行 for(int i = n - pos - 1; i >= pos; i--) result[n-1-pos][i] = count++; //左边一列 for (int i = n - pos - 2; i > pos; i--) result[i][pos] = count++; if (count <= n * n) Matrix(n, pos+1, count, result); return; } vector<vector<int>> generateMatrix(int n) { vector<vector<int>> result; if (n <= 0) return result; else if (n == 1) { result.push_back(vector<int> (1,1)); return result; } else { for(int i=0;i<n;i++) { result.push_back(vector<int> (n, 0)); } Matrix(n, 0, 1, result); } return result; }};
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