1.合并有序数组
方法一：清空符合条件的原数组
function merge(l1,l2){
let res = [];
while(l1.length && l2.length){
if(l1[0]<l2[0]){
res.push(l1.shift())
}else{
res.push(l2.shift())
}
}
res = res.concat(l1,l2);
return res
}
方法二：双指针
function merge(l1,l2){
let res = [];
let i=0;j=0
while(i<l1.length && j<l2.length){
if(l1[i]<l2[j]){
res.push(l1[i])
i++;
}else{
res.push(l2[j])
j++
}
}
while(i<l1.length){
res.push(l1[i])
i++;
}
while(j<l2.length){
res.push(l2[j])
j++;
}
return res
}
let l1 = [1,4,5,23];
let l2 = [0,2,3,6,7,8];
console.log(merge(l1,l2))
2.用指定数组内容按照数字顺序替换占位符@1@，@2@
例子：str = 你好@1@，天气@2@ arr = ['李明','晴'] =》你好李明，天气晴
function cusReplce(str,repArr){
for(let i=0; i<repArr.length; i++){
str = str.replace(new RegExp('@'+(i+1)+'@'),repArr[i])
}
return str;
}
let str = '你好@1@，天气@2@';
let arr = ['李明','晴朗'];
console.log(cusReplce(str,arr))
3.多维数组扁平化
function flatter(list){
if(Array.isArray(list)){
return list.reduce((a,b)=>{
return [...a, ...flatter(b)]
},[])
}
return [list];
}
let arr = [1,2,[3,[4,5]]];
console.log(flatter(arr))
4.M*N的棋盘从左上角到右下角总共有多少种走法,要求不能回头
function chess(m,n){
if(m<1||n<1) return 0;
if(m==1 && n!==1){
return n+1;
}
if(m!=1 && n==1){
return m+1;
}
return chess(m-1,n)+chess(m,n-1)
}
5.二叉树是否存在路径总和等于目标值
function path(root,target){
if(!root){
return false
}
if(!root.left&&!root.right){
return root.value == target;
}
return path(root.left,target-root.value) || path(root.right,target-root.value);
}
6.var locationList = [
{ id: 0, name: "中国" },
{ id: 1, pid: 0, name: "广东省" },
{ id: 2, pid: 1, name: "深圳市" },
{ id: 3, pid: 1, name: "广州市" },
{ id: 4, pid: 0, name: "陕西省" },
]
变为
[{
id: 0,
name: "中国" ,
children:[
{ id: 1,
pid: 0,
name: "广东省",
children:
[
{ id: 2, pid: 1, name: "深圳市" ,children:[]},
{ id: 3, pid: 1, name: "广州市",children:[] }
]
},
{ id: 4, pid: 0, name: "陕西省" ,children:[]},
]
}].
function buildLocationTree(list){
let map = {};
for(let i=0; i<list.length;i++){
let id = list[i]['id'];
map[id] = list[i]
}
let res = [];
for(let i=0; i<list.length;i++){
let item = list[i];
let pid = item['pid'];
if(pid !== undefined && map[pid]){
if(map[pid].children){
map[pid].children.push(item)
}else{
map[pid].children = [item]
}
}else{
res.push(item)
}
}
return res;
}
7.树的遍历
let tree = [{"id":0,"name":"中国","children":[{"id":1,"pid":0,"name":"广东省","children":[{"id":2,"pid":1,"name":"深圳市"},{"id":3,"pid":1,"name":"广州市"}]},{"id":4,"pid":0,"name":"陕西省"}]}]
找出指定id的name
function dfs(list,id){
for(let i=0;i<list.length;i++){
let item = list[i];
if(item.id == id){
return item.name;
}
else if(item.children && item.children.length>0){
dfs(item.children,id)
}
else{
return -1
}
}
}
function find(lists,id){
if(lists.id == id){
return lists.name;
}
else{
if(lists.children && lists.children.length>0){
return dfs(lists.children,id)
}
}
}
console.log(find(tree,6))
8.add(1,2)(2)() // 5
function add(){
let args = [...arguments];
let res = 0;
for(let i = 0; i<args.length;i++){
res += args[i];
}
return function F(...args1){
if(args1.length){
return add(...args,...args1)
}
else{
return res;
}
}
}
console.log(add(1,2)(4)())
9.实现一个console和setTimeout链式调用
class A{
constructor(){
this.promise = Promise.resolve()
}
console(v){
this.promise = this.promise.then(()=>{
console.log(v)
})
return this;
}
setTimeout(wait){
this.promise = this.promise.then(()=>{
return new Promise((resolve)=>{
setTimeout(()=>{
resolve()
},wait)
})
})
return this;
}
}
let a = new A()
a.console('1').setTimeout(1000).console('2')
10.实现一个sleep
function sleep(wait){
return new Promise((resolve,reject)=>{
setTimeout(resolve,wait)
})
}
async function init(wait){
let res = await sleep(wait)
console.log(3)
return res;
}
console.log(1)
init(3000)
11.实现compose函数
function compose(...args){
let len = args.length;
let count = len-1;
let res = '';
return function F(...args1){
res = args[count].apply(this,args1);
if(count <= 0){
return res
}else{
count--;
return F(res)
}
}
}
let test = (x,y) => x+y
let uppcase = (x) => x.toUpperCase()
let add = (x) => x+'123'
let re = compose(add, uppcase,test)('hello','world')
console.log(re)
- fn([['a', 'b'], ['n', 'm'], ['0', '1']]) => ['an0', 'am0', 'an1', 'am1', 'bn0', 'bm0', 'bn1', 'bm0']
function dfs(result, tempRes, curIndex, list){
if(tempRes.length === list.length){
result.push(tempRes.join(''));
return;
}
for(let i=0;i<list[curIndex].length;i++){
tempRes.push(list[curIndex][i]);
dfs(result,tempRes,curIndex+1,list);
tempRes.pop()
}
}
function composeList(list){
let len = list.length;
let result = [];
dfs(result,[],0,list);
return result;
}
let term = composeList([['a', 'b'], ['n', 'm'], ['0', '1']])
console.log(term)
13.给数组中的字符串编号，f(['ab', 'c', 'd', 'ab', 'c']) => ['ab1', 'c1', 'd', 'ab2', 'c2']
function fn(list){
let map = {};
for(let i=0;i<list.length;i++){
if(map[list[i]]){
map[list[i]]++;
}
else{
map[list[i]] = 1;
}
list[i] = list[i]+map[list[i]]
}
return list;
}
let list = fn(['ab', 'c', 'd', 'ab', 'c'])
console.log(list)
- 判断链表有环
1.快慢双指针 相遇则有环
function circle(head){
let p1 = head;
let p2 = head;
while(p2 && p2.next){
p1 = p1.next;
p2 = p2.next.next
if(p1 === p2){
return true;
}
}
return false;
}
2.使用set判断节点有没有加入过，有的话则有环
function circle(head){
let set = new Set();
while(head){
if(set.has(head)) return true
set.add(head)
head = head.next;
}
return false;
}
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