javascript二维数组的面试题

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]))

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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;}

登录后复制由于queue只是临时存匹配结果的，还要出队列进行新一轮的匹配所以只要再用一个缓存队列存储过往匹配的成功数据即可，没有必要最后在进行遍历的必要。代码如下：

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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