如何在不改变原始数组的情况下从时间复杂度为O(n)或更好的排序数组中获取唯一值
我想在不改变原始数组的情况下计算给定数组中的唯一值,但解决方案必须在time complexity of O(n). 到目前为止,所有我见过的解决方案,有time complexity of O(n^2) 喜欢这里。我在解决方案的逻辑中找不到错误。我是数据结构和算法的新手,想要一个简单的解决方案。
我的代码 -
const countUniqueValues = (arr) =>{
if(arr.length === 0){
return console.log(arr.length);
}else if(arr.length === 1){
return console.log(arr.length);
}
const unique = [];
let i = 0;
for( let j = 1; j < arr.length; j++){
if(arr[i] !== arr[j]){
i ++;
unique.push(arr[i]);
}
}
return console.log(unique);
}
//test cases
countUniqueValues([1,1,1,1,1,2]) // 2
countUniqueValues([1,2,3,4,4,4,7,7,12,12,13]) // 7
countUniqueValues([]) // 0
countUniqueValues([-2,-1,-1,0,1]) // 4
错误的输出 -
[ 1 ]
[
2, 3, 4, 4,
4, 7, 7, 12
]
0
[ -1, -1, 0 ]
回答
将数组转换为 Set ( O(n)) 并计算集合的大小:
const countUniqueValues = arr => new Set(arr).size;
- If the interviewer rejects this, the shortest and most easily understandable solution, I would not want to work at such a company. Good, professional, maintainable code is **short**, **simple**, and easy to understand at a glance; it's not over-engineered just for the heck of it.
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