前端算法面试题汇总

数组去重

function unique(arr) {    const seen=new Map();    return arr.filter((a) => !seen.has(a)&& seen.set(a,1));}//orfunction unique(arr) {    return Array.from(new Set(arr))}//or[...(new Set(arr))];

求mul(1)(2)(3),其中调用次数未限定

单调用三次

var mul = function(a){    return function(b){        return function(c){            return a*b*c;        };    };};

function mul(x) {    var product= x;    var tmp = function (y) {        product = product * y;        return tmp;    };    tmp.toString = function () {        return product ;    };    return tmp;}console.log(mul(1)(2)(3));  //6console.log(mul(1)(2)(3)(4));   //24

查找两个节点的最近公共父节点

function commonParentNode(oNode1,oNode2) {    var nodes1=[],nodes2=[];    while (oNode1){        nodes1.push(oNode1);        oNode1=oNode1.parentNode;    }    while (oNode2){        nodes2.push(oNode2);        oNode2=oNode2.parentNode;    }    while (( a = nodes1.pop()) === nodes2.pop()){        node = a;    }    return node;}

function commonParentNode(oNode1,oNode2) {    if(!oNode1||!oNode2){        return null;    }    for(;oNode1;oNode1=oNode1.parentNode){        if(oNode1.contains(oNode2)){            return oNode1;        }    }}

求1000以内的质数

function prime(n) {    var arr=[];    for(var i=2;i<n;i++){        for(var j=2;j<Math.sqrt(i);j++){            if(i%j===0){                break;            }        }        if(j>=Math.sqrt(i)){            arr.push(i);        }    }    return arr;}