岛数二维数组算法 [英] island count two-dimensional array algorithm

问题描述

let arr = [[1, 0, 1],
           [1, 0, 0],
           [1, 1, 1]
     ];

我有一个数组，其中 1 个岛和 0 个水.我需要写一个岛屿计数器.这里有 2 个岛 1 个大和 1 个小(单个).例如这里有 5 个单岛

i have array where 1-island and 0-water. I need to write a island counter. Here are 2 island 1 big and 1 small(singele). For example here are 5 single islands

let arr = [[1, 0, 1],
           [0, 1, 0],
           [1, 0, 1]
     ]; 

我写了双循环来上诉数组中的每个项目，如下所示:

i had write double cycle to appeal each item in array like this:

for(let i = 0; i < arr.length; i++){
    for(let x = 0; x < arr[i].length; x++){
         if(...){}
    }
 }

我需要为此写一个条件.请帮帮我.

and I need to write сondition for this. Help me please.

推荐答案

您可以使用计数器并检查所有相邻项目并使用实际计数器更新元素.

You could use a counter and check all adjacent items and update the element with the actual counter.

function check(array) {

    function test(array, i, j, value) {
        if (array[i] && array[i][j] === -1) {
            array[i][j] = value;
            test(array, i -1, j, value);
            test(array, i + 1, j, value);
            test(array, i, j - 1, value);
            test(array, i, j + 1, value);
            return true;
        }
    }
    var value = 1;

    array.forEach(a=> a.forEach((b, i, bb) => bb[i] = -b));
    array.forEach((a, i, aa) => a.forEach((b, j) => test(aa, i, j, value) && value++));
    document.getElementById('out').innerHTML += array.map(a => a.join(' ')).join('\n') + '<hr>';
    return value - 1;
}

console.log(check([[1, 0, 1], [1, 0, 0], [1, 1, 1]]));
console.log(check([[1, 0, 1], [0, 1, 0], [1, 0, 1]]));

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