LeetCode - Max Area of Island (Javascript)

Description

위에서 풀었던 문제와 굉장히 유사하다. 그냥 그 문제에서 최대 넓이만 구하면 되는 문제다!

You are given an m x n binary matrix grid. An island is a group of 1‘s (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

The area of an island is the number of cells with a value 1 in the island.

Return the maximum area of an island in grid. If there is no island, return 0.

Example 1:

Input: grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1,0,0],[0,1,0,0,1,1,0,0,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,0,1,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,0,1,1,0,0,0,0]]
Output: 6
Explanation: The answer is not 11, because the island must be connected 4-directionally.

Example 2:

Input: grid = [[0,0,0,0,0,0,0,0]]
Output: 0

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 50
• grid[i][j] is either 0 or 1.

---

소스 코드

function maxAreaOfIsland(grid) {
  let answer = 0;
  let count = 0;
  const maxRow = grid.length - 1;
  const maxColumn = grid[0].length - 1;

  function traverse(row, col, grid) {
    // 아래 조건에 해당되면 해당 노드의 값은 1이 아닌 것!
    if (
      grid[row] === undefined ||
      grid[row][col] === undefined ||
      grid[row][col] === 0
    ) {
      return;
    }
    grid[row][col] = 0;
    // 매 재귀 실행 시, 1을 찾을 때마다 count++을 해준다.
    count++;
    traverse(row - 1, col, grid);
    traverse(row, col + 1, grid);
    traverse(row + 1, col, grid);
    traverse(row, col - 1, grid);
  }

  for (let i = 0; i <= maxRow; i++) {
    for (let j = 0; j <= maxColumn; j++) {
      if (grid[i][j] === 1) {
        traverse(i, j, grid);
        // 그리고 재귀가 끝났을 때, answer와 count 중 더 큰 값을 answer에 할당해준다.
        answer = Math.max(answer, count);
        // 그리고 다시 count는 0으로 초기화 해주면 된다.
        count = 0;
      }
    }
  }
  return answer;
}