#90 判断两个 Set 是否相同

[ScriptOJ]

完成 isSameSet 函数，它接受了两个 Set 对象作为参数，请你返回 true/false 来表明这两个 set 的内容是否完全一致，例如：

const a = {}
const b = 1
const c = 'ScriptOJ'

const set1 = new Set([a, b, c])
const set2 = new Set([a, c, b])

isSameSet(set1, set2) // => true

[陈小俊]

const isSameSet = (set1, set2) => [...set1].every((o) => set2.has(o)) && [...set2].every((o) => set1.has(o))

[ibufu]

const isSameSet = (s1, s2) => {
  if (s1.size !== s2.size) {
    return false;
  }
  return [...s1].every(i => s2.has(i))
}

[Reson_a]

const isSameSet = (set1, set2) => {
    let s = new Set([...set1, ...set2])
    return s.size == set1.size && s.size == set2.size
}

[binnear]

const isSameSet = (s1, s2) => s1.size === s2.size && [...s1].filter(x => s2.has(x)).length === s1.size;

[rony-land]

const isSameSet = (s1, s2) => {
for(let item of s1){
if(s2.has(item)){
s2.delete(item);
}else{
return false;
}
}
if(s2.size == 0){
return true;
}else{
return false;
}
}

[zhouyu]

const isSameSet = (s1, s2) => {
  function isContain(aSet, bSet){
    return ![...aSet].some(item => !bSet.has(item))
  }
  return isContain(s1, s2) && isContain(s2, s1);
}

[zyz565822]

const isSameSet = (s1, s2) => {
const getComp = v=>JSON.stringify([...v].sort());
return getComp(s1) === getComp(s2)
}

[fanxin2008]

@陈小俊 此处的has用法是哪个对象的方法？

[fanxin2008]

@陈小俊 我看懂了~

[Tvinsh]

const isSameSet = (s1, s2) => {
let diff1 = new Set([...s1].filter(x => !s2.has(x)))
let diff2 = new Set([...s2].filter(x => !s1.has(x)))
if(diff1.size || diff2.size) {
return false
}
return true
}

[愤怒地小跑]

写的不好，但我能想到的就是这样了

const isSameSet = (s1, s2) => {

   function fn(...rest) {
       let arr = [];
       for (let i = 0; i < rest.length; i++) {
           for (let k of rest[i]) {
               arr.push(k);
           }
       }
       return new Set(arr);
   }

   if (s1.size === s2.size) {
       if (fn(s1, s2).size === s1.size) {
           return true;
       }
       return false;
   }
   return false;
};

[clf_programing]

@Reson_a 这个方法使用了set的不重复属性，如果两个set相等，合并后和合并前的size一定相等

[Jervis]

		const isSameSet = (s1, s2) => {/* TODO */

			let arr1 = [...s1] , arr2 = [...s2];

			let a = arr1.every(item => arr2.includes(item))
			let b = arr2.every(item => arr1.includes(item))

			return a && b && arr1.length == arr1.length
	}

[卡车拉萝卜司机]

const isSameSet = (s1, s2) => /* TODO */{
var resArr = new Set([]);
for(let item of s1){
resArr.add(s2.has(item))
}
return !resArr.has(false)
}
为什么这样不可以过？

[Embrace]

const isSameSet = (s1, s2) => {
if(s1.size !== s2.size){return false}
let arr1 = [...s1];
let arr2 = [...s2];
let set = new Set([...arr1, ...arr2]);
let arr3 = Array.from(set);
return arr3.length === arr2.length
}

[SimonMa]

[corazon]

const isSameSet = (s1, s2) => {
  return (s1.size==s2.size && new Set([...s1,...s2]).size == s1.size)?true:false
}

[liang5982]

@Reson_a 简洁，实用，好办法