54. Spiral Matrix && 59. Spiral Matrix II

Posted 2020-11-24 qiang-wei

Given a positive integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.

Example:

Input: 3
Output:
[
 [ 1, 2, 3 ],
 [ 8, 9, 4 ],
 [ 7, 6, 5 ]
]

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {

        vector<vector<int>> res(n,vector<int>(n,0));
        if (n == 0) return res;
        int i = 1;
        int rowS = 0,rowE = n - 1,colS = 0,colE = n -1;
        while(i <= n * n)
        {
            for (int j = colS;j <= colE;++j)
            {
                res[rowS][j] = i++;
            }
            rowS++;
            
            for (int j = rowS;j <= rowE;++j)
            {
                res[j][colE] = i++;
            }
            colE--;
            
            for (int j = colE;j >= colS;--j)
            {
                res[rowE][j] = i++;
            }
            rowE--;
            
            for (int j = rowE;j >= rowS;--j)
            {
                res[j][colS] = i++;
            }
            colS++;
        }
        return res;
    }
};

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

Example 1:

Input:
[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]

Example 2:

Input:
[
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9,10,11,12]
]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = m ? matrix[0].size() : 0, u = 0, d = m - 1, l = 0, r = n - 1, p = 0;
        vector<int> order(m * n);
        while (u <= d && l <= r) {
            for (int col = l; col <= r; col++) {
                order[p++] = matrix[u][col];
            }
            if (++u > d) {
                break;
            }
            for (int row = u; row <= d; row++) {
                order[p++] = matrix[row][r];
            }
            if (--r < l) {
                break;
            }
            for (int col = r; col >= l; col--) {
                order[p++] = matrix[d][col];
            }
            if (--d < u) {
                break;
            }
            for (int row = d; row >= u; row--) {
                order[p++] = matrix[row][l];
            }
            if (l++ > r) {
                break;
            }
        }
        return order;
    }
};