字符串出现的不重复最长长度

昨日回顾

1. 归并排序

// 1. reduce
function a1(arr){
  return arr.reduce((acc, cur)=>{
    return Array.isArray(cur)? [...acc, ...a1(cur))]:[...acc, cur];
  },[])
}
// 2. concat
function a2(arr){
  let arrCopy = [...arr];
  while(arrCopy.some(item=>Array.isArray(item))){
    arrCopy = [].concat([...arrCopy]);
  }
  return arrCopy;
}
// 3. flat Infinity
function a3(arr){
  return arr.flat(Infinity).sort((a-b)=>a-b);
}
// 4. merge mergeSort reduce
function a4(arr){
  function  flat(aa){
    return aa.reduce((acc, cur)=>{
      return Array.isArray(cur)? [...acc, ...flat(cur)]:[...acc, cur];
    },[]);
  }
  function mergeSort(arrS){
    if(arrS.length<2) return arrS;
    let middle = arrS>>1;
    let left = arrS.slice(0, middle);
    let right = arrS.slice(middle);
   return  merge(mergeSort(left), mergeSort(right));
  }
  function merge(left, right){
    let arrC;
    while(left.length && right.length){
      if(left.shift()<right.shift()){
        arrC.push(left.shift());
      } else{
        arrC.push(right.shift());
      }
    }
    while(left.length){
      accC.push(left.shift())
    }
    while(right.length){
      accC.push(right.shift())
    }
    return accC;
  }
  return mergeSort(flat(arr));
}

2. 斐波那契

function fibonacci1(n, v1 = 1, v2 = 1) {
  if (n == 1) return v1;
  if (n == 2) return v2;
  return fibonacci1(n - 1, v2, v1 + v2);
}
function fibonacci2(n) {
  let pre = 1;
  let cur =1;
  let data;
  if(n==1 || n==2) return 1;
  for(let k =3; k<=n; k++){
    data = pre+cur;
    pre=cur;
    cur=data;
  }
  return data;
}
function

今日解题

let str = "fdsfsdzffdfdddfsdsds";

let arr = [];
const s = str.split("");
let total = 0; // 长度
let maxStr = ""; // 最长不重复长度的字符串
for (let i = 0; i < s.length; i++) {
  const ele = s[i];
  const idx = arr.indexOf(ele);
  if (idx > -1) {
    arr = arr.slice(idx + 1);
    console.log(arr);
  }
  arr.push(ele);
  if (arr.length > total) {
    maxStr = arr.join("");
    total = arr.length;
  }
}
console.log(total, maxStr);