Leetcode 54: spiral matrix / Leetcode59: spiral matrixII

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].

对于螺旋打印一个matrix，可以考虑一圈一圈打印，从外圈到内圈是用offset来传递。对于base case 情况的考虑，matrix可能没有元素，或者可能只有一列或者一行
时间复杂度： O（n^2）空间复杂度： O（n）

 class Solution {
private:  
    void helper(vector<int> &rst, vector<vector<int>>& matrix, int row, int col, int offset) {
        //base case
        if (row <= 0 || col <= 0) return;
        if (row == 1 ) {
            for (int i = 0; i < col; i++) {
                rst.push_back(matrix[offset][i+offset]);
            }
             return;
        }
        if (col == 1) {
            for (int i = 0; i < row; i++) {
                rst.push_back(matrix[offset+i][offset]);
            }
            return;
        }
        //from left to right
        for (int i = 0; i < col-1; i++) {
            rst.push_back(matrix[offset][offset+i]);
        }
        //from up to down
        for (int i = 0; i < row-1; i++) {
            rst.push_back(matrix[offset+i][col-1+offset]);
        }
        //from right to left
        for (int i = col-1; i >= 1; i--) {
            rst.push_back(matrix[row-1+offset][offset + i]);
        }
        //from down to up
        for (int i = row - 1; i >= 1; i--) {
            rst.push_back(matrix[i + offset][offset]);
        }
        helper(rst, matrix, row - 2, col - 2, offset + 1);
    }
public:
//use recursion
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int> rst;
        //sanity check
        if (matrix.size() == 0 || matrix[0].size() == 0) return rst;
        helper(rst, matrix, matrix.size(), matrix[0].size(), 0);
        return rst;
    }
};

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 ]
] 

解法与上题类似， 初始化一个二维矩阵：  vector<int> solu(n,0); vector<vector<int>> rst(n,solu);

class Solution {
private:
    void helper( int n, int count , vector<vector<int>> & matrix, int offset){
        //base case
        if (n == 0) return;
        if (n == 1) {
            matrix[offset][offset] = count;
            return;
        }
        //from left to right
        for (int i = 0; i < n-1; i++) {
            matrix[offset][i+offset] = count++;
        }
        //from up to down
        for (int i = 0; i < n-1; i++) {
            matrix[offset + i][n-1+offset] = count++;
        }
        //from right to left
        for (int i = n-1; i >= 1; i--) {
            matrix[offset+ n-1][offset+i] = count++;
        }
        //from down to up
        for (int i = n-1; i >= 1; i--) {
            matrix[offset+i][offset] = count++;
        }
        helper(n-2, count, matrix, offset+1);
    }
public:
    vector<vector<int>> generateMatrix(int n) {
        vector<int> solu(n,0);
        vector<vector<int>> rst(n, solu); 
        //sanity check
        if (n == 0) return rst;
        helper(n, 1, rst, 0);
        return rst;
    }
};