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:
Input: grid = [ ["1","1","1","1","0"], ["1","1","0","1","0"], ["1","1","0","0","0"], ["0","0","0","0","0"] ] Output: 1
Example 2:
Input: grid = [ ["1","1","0","0","0"], ["1","1","0","0","0"], ["0","0","1","0","0"], ["0","0","0","1","1"] ] Output: 3
Constraints:
m == grid.lengthn == grid[i].length1 <= m, n <= 300grid[i][j]is'0'or'1'.
Idea:
Traverse connected components: Use DFS to traverse 4 directions and mark them “0” if found “1”
Solution:
/**
* @param {character[][]} grid
* @return {number}
*/
var numIslands = function(grid) {
let row = grid.length;
if (row === 0) return 0;
let col = grid[0].length;
let res = 0;
for (let x = 0; x < col ; x++) {
for (let y = 0; y < row; y++) {
if (grid[y][x] === "1") {
res++;
dfs(grid, x, y, col, row);
}
}
}
return res;
};
var dfs = function(grid, x, y, n, m) {
// over boundry cases and not "1" => return
if (x < 0 || y < 0 || x >= n || y >= m || grid[y][x] === "0") return;
// already visited mark it "0"
grid[y][x] = "0";
// traverse 4 directions if any "1" => mark them "0"
dfs(grid, x + 1, y, n, m);
dfs(grid, x - 1, y, n, m);
dfs(grid, x, y + 1, n, m);
dfs(grid, x, y - 1, n, m);
}