[JS] Problem solving | #Number of islands#

Problem solving | #Number of islands#

topic link

number of islands

---

Topic description

Time limit: 1 second Space limit: 256M

describe

Given a 01 matrix, 1 represents land, 0 represents ocean, if two 1s are adjacent, then the two 1s belong to the same island. We only consider up, down, left and right as adjacent.
Island: Adjacent land can form an island (adjacent: up, down, left and right) Determine the number of islands.

E.g:

enter

[
[1,1,0,0,0],
[0,1,0,1,1],
[0,0,0,1,1],
[0,0,0,0,0],
[0,0,1,1,1]
]

The corresponding output is

3

Example 1

enter:

[[1,1,0,0,0],[0,1,0,1,1],[0,0,0,1,1],[0,0,0,0,0],[0,0,1,1,1]]

return value:

3

Example 2

enter:

[[0]]

return value:

0

Example 3

enter:

[[1,1],[1,1]]

return value:

1

Remark:

01 Matrix range <=200*200

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Problem solving ideas

A very typical dyeing method, perform dfs on each point, fully dye the adjacent points with a value of 1, and use sum to record the current dyeing value.

In short, the first island is fully stained as -1, the second island is fully stained as -2, and finally the inverse of the largest negative number is the correct answer.

Using a dfs will do the trick.

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problem solving code

/**
 * 判断岛屿数量
 * @param grid string字符串型二维数组 
 * @return int整型
 */
function solve(grid) {
    
    
    // write code here
    // grid = [[1, 1, 0, 0, 0], [0, 1, 0, 1, 1], [0, 0, 0, 1, 1], [0, 0, 0, 0, 0], [0, 0, 1, 1, 1]];
    let col = grid.length;
    let row = grid[0].length;
    let sum = 0;
    let dfs = (x, y) => {
    
    
        if (grid[x][y] == 1) {
    
    
            grid[x][y] = sum;
        }
        let tox = x + 1;
        let toy = y;
        if (tox >= 0 && tox < col && toy >= 0 && toy < row && grid[tox][toy] == 1) {
    
    
            dfs(tox, toy);
        }
        tox = x - 1;
        toy = y;
        if (tox >= 0 && tox < col && toy >= 0 && toy < row && grid[tox][toy] == 1) {
    
    
            dfs(tox, toy);
        }
        tox = x;
        toy = y + 1;
        if (tox >= 0 && tox < col && toy >= 0 && toy < row && grid[tox][toy] == 1) {
    
    
            dfs(tox, toy);
        }
        tox = x;
        toy = y - 1;
        if (tox >= 0 && tox < col && toy >= 0 && toy < row && grid[tox][toy] == 1) {
    
    
            dfs(tox, toy);
        }
        return;
    }
    for (let i = 0; i < col; i++) {
    
    
        for (let j = 0; j < row; j++) {
    
    
            if (grid[i][j] == 1) {
    
    
                sum--;
                dfs(i, j);
            }
        }
    }
    sum = sum < 0 ? -sum : sum;
    return sum;
}