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코딩테스트 스터디
코딩테스트 스터디/
Number of Islands

Number of Islands

문제

Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.
An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
Example 1:
Example 2:
Constraints:
  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 300
  • grid[i][j] is '0' or '1'.

풀이

은찬
재영
효성
Link
https://leetcode.com/problems/number-of-islands/
Deadline
May 15, 2022
Status
Archived
Type
DFS
BFS
union-find
Matrix
Input: grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
Output: 1

Input: grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
Output: 3

const numIslands = (grid) => {
    const dx = [-1, 1, 0, 0];
    const dy = [0, 0, -1, 1];
    const m = grid.length;
    const n = grid[0].length;
    const visited = Array.from({length: m}, () => Array(n).fill(false));
    let answer = 0;
    
    const checkRange = (x, y) => {
        if(x >= 0 && x < m && y >= 0 && y < n){
            return true;
        }
        
        return false;
    }
    
    const dfs = (x, y) => {
        for(let i = 0; i < 4; i++){
            const nx = x + dx[i];
            const ny = y + dy[i];
            
            if(checkRange(nx, ny) && !visited[nx][ny] && grid[nx][ny] == "1"){
                visited[nx][ny] = true;
                dfs(nx, ny);
            }
        }
    }
    
    for(let i = 0; i < m; i++){
        for(let j = 0; j < n; j++){
            if(grid[i][j] == "1" && !visited[i][j]){
                visited[i][j] = true;
                dfs(i, j);
                answer++;
            }
        }
    }
    
    return answer;
};
const numIslands = (grid) => {
  const bfs = (x, y) => {
    const directions = [
      [-1, 0],
      [1, 0],
      [0, -1],
      [0, 1],
    ];

    const queue = [];
    queue.push([x, y]);

    while (queue.length) {
      const [x, y] = queue.shift();
      if (!grid[x][y]) continue;
      grid[x][y] = 0;

      for (let i = 0; i < 4; i += 1) {
        const [dx, dy] = directions[i];

        const nx = x + dx;
        const ny = y + dy;

        if (
          nx >= 0 &&
          nx < rowLength &&
          ny >= 0 &&
          ny < colLength &&
          grid[nx][ny] === '1'
        ) {
          queue.push([nx, ny]);
        }
      }
    }
  };

  let isLandCount = 0;

  const rowLength = grid.length;
  const colLength = grid[0].length;

  for (let i = 0; i < rowLength; i += 1) {
    for (let j = 0; j < colLength; j += 1) {
      const now = grid[i][j];

      if (now === '1') {
        bfs(i, j);
        isLandCount += 1;
      }
    }
  }

  return isLandCount;
};
var numIslands = function(grid) {
    const m = grid.length;
    const n = grid[0].length;
    
    let answer = 0;
    
    const canGo = (x, y) => {
        return x < m && y < n && x >= 0 && y >= 0;
    }
    
    const dfs = (x, y) => {
        if(!canGo(x,y) || grid[x][y] === '0') {
            return;
        }
        grid[x][y] = '0';
        dfs(x + 1, y);
        dfs(x - 1, y);
        dfs(x, y + 1);
        dfs(x, y - 1);
    }
    
    for(let i = 0; i < m; i++) {
        for(let j = 0; j < n; j++) {
            if(grid[i][j] === '1') {
                answer++
                dfs(i, j);
            }
        }
    }
    
    return answer;
};