javascript二维数组的面试题

本文主要和大家分享一个关于javascript二维数组的面试题，希望能帮助到大家。给定一个二维数组，实现一个功能函数 fn，向这个函数中传递这个二维数组的一个坐标，如果这个坐标的值为 ”1“，将返回和这个坐标所有相连的并且坐标值为1坐标。

例如,传递了 fn([3,4])得到的结果为：
[[3,4],[4,4],[5,4],[6,4],[7,4],[8,4],[8,5],[8,6]]

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

解答思路：宽度优先遍历即可。供参考，相连条件没给出且当做是横竖方向:

var arr =[ 
[0,0,0,0,0,0,0,0,0,0,0], 
[0,0,0,0,0,0,0,0,0,0,0], 
[0,0,0,0,0,0,0,0,0,0,0], 
[0,0,0,0,1,0,0,0,1,0,0], 
[0,0,0,0,1,0,0,0,1,0,0],
[0,0,0,0,1,0,0,0,1,0,0],
[0,0,0,0,1,0,0,0,0,0,0],
[0,0,0,0,1,0,0,0,0,0,0],
[0,0,0,0,1,1,1,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
] 

function fn ([x, y]) {
    if (arr[x][y] !== 1) return false
    
    const queue = [[x, y]]
    const memo = arr.map(row => new Array(row.length).fill(false))
    const direction = [
     [-1, 0],
     [1, 0],
     [0, -1],
     [0, 1],
    ]
    
    while(queue.length > 0) {
        const [x, y] = queue.pop()
        direction.forEach(([h, v]) => {
            const newX = x + h
            const newY = y + v
            if (arr[newX][newY] === 1 && !memo[newX][newY]) {
                memo[newX][newY] = true
                queue.push([newX, newY])
            }
        })
    }
    
    const result = []
    for (let i = 0; i < arr.length; i++) {
        for (let j = 0; j < arr[i].length; j++) {
            if(memo[i][j]) {
                result.push([i, j])
            }
        }
    }
    return result
}

console.log(fn([3,4]))

借鉴前两位 对象+递归实现

function fn(point) {
    var memo = {}, result = [], direction = [[-1, 0],[1, 0],[0, -1],[0, 1]]
    function dg([x, y]) {
        result.push(memo[x + "," + y] = [x, y]);
        direction.forEach(([h, v]) => {
            const newX = x + h
            const newY = y + v
            if (arr[newX][newY] === 1 && !memo[newX + "," + newY]) {
                dg([newX, newY]);
            }
        })
    }
    dg(point);
    return result;
}

var arr =[

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

function fn ([x, y]) {
    if (arr[x][y] !== 1) return false
    
    const queue = [[x, y]]
    const result = [[x,y]]
    const memo = arr.map(row => new Array(row.length).fill(false))
    const direction = [
     [-1, 0],
     [1, 0],
     [0, -1],
     [0, 1],
    ]
    
    while(queue.length > 0) {
        const [x, y] = queue.pop()
        direction.forEach(([h, v]) => {
            const newX = x + h
            const newY = y + v
            if (arr[newX][newY] === 1 && !memo[newX][newY]) {
                memo[newX][newY] = true
                queue.push([newX, newY])
                result.push([newX, newY])
            }
        })
    }
    
    return result
}

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