README.md

Spiral Matrix

Question:

Given an m x n matrix, return all elements of the matrix in spiral order. (See example graph on LeetCode for better visualization).

How to Solve:

This question isn't tough per se -- it's just tedious to keep track of the indices.

We initialize four variables respectively pointing to the row begin, column begin, row end, and column end. The former two starts at zero an increases, while the latter two starts at height-1 and width-1 and decreases.

We then wrap four for-loops inside a big while-loop, each time incrementing row being and column begin, while decreasing row end and column end, until row begin surpasses row end, or column begin surpasses column end.

Having this in mind, the code is rather self-explanatory.

My C++ Solution:

class Solution {
 public:
  vector<int> spiralOrder(vector<vector<int>> &mat) {
    if (mat.size() == 0) return {};
    // Initialize boundaries
    int row_begin = 0, row_end = mat.size() - 1;
    int col_begin = 0, col_end = mat[0].size() - 1;
    vector<int> ans;
    while (row_begin <= row_end && col_begin <= col_end) {
      for (int i = col_begin; i <= col_end; ++i) {
        ans.push_back(mat[row_begin][i]);
      }
      ++row_begin;

      for (int i = row_begin; i <= row_end; ++i) {
        ans.push_back(mat[i][col_end]);
      }
      --col_end;

      if (row_begin <= row_end) {  // still rows to traverse through
        for (int i = col_end; i >= col_begin; --i) {
          ans.push_back(mat[row_end][i]);
        }
        --row_end;
      }

      if (col_begin <= col_end) {  // still cols to traverse through
        for (int i = row_end; i >= row_begin; --i) {
          ans.push_back(mat[i][col_begin]);
        }
        ++col_begin;
      }
    }
    return ans;
  }
};