this is what I do now. Is there a better way?
let flag = true
const listA = [1, 2, 3]
const listB = [2, 3, 4]
if (listA.length !== listB.length) {
flag = false
} else {
listA.forEach(item => {
if (listB.indexOf(item) === -1) {
flag = false
}
})
}it can be done in one sentence:
const checkArrSames = (arr1, arr2) => [...new Set(arr1)].length === [...new Set([...arr1, ...arr2])].length;A solution that conforms to the meaning of the question function unique(arr) { const res = new Map(); return arr.filter((a) => !res.has(a) && res.set(a, 1)) } because map key is unique, you can change the an in res.has (a) to whate...
for example, duplicates according to the name attribute in the an array element. is this the simplest way? function unique(arr) { const res = new Map(); return arr.filter((a) => !res.has(a.name) && res.set(a.name, 1)) } reference...
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