HackMD
1207.Unique Number of Occurrences
tags: Easy,Array,Hash Table
1207. Unique Number of Occurrences
題目描述
Given an array of integers arr, return true if the number of occurrences of each value in the array is unique, or false otherwise.
範例
Example 1:
Input: arr = [1,2,2,1,1,3]
Output: true
Explanation: The value 1 has 3 occurrences, 2 has 2 and 3 has 1. No two values have the same number of occurrences.
Example 2:
Input: arr = [1,2]
Output: false
Example 3:
Input: arr = [-3,0,1,-3,1,1,1,-3,10,0]
Output: true
Constraints:
1 <= arr.length <= 1000-1000 <= arr[i] <= 1000
解答
Java
class Solution {
public boolean uniqueOccurrences(int[] arr) {
HashMap<Integer, Integer> hm = new HashMap<>();
for(Integer num : arr) {
hm.put(num, hm.getOrDefault(num, 0) + 1);
}
return hm.size() == new HashSet<>(hm.values()).size();
}
}
KobeWed, Nov 30
Python
class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
return len(Counter(arr)) == len(set(Counter(arr).values()))
KobeWed, Nov 30
class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
mapping = {}
for v in arr:
if v not in mapping:
mapping[v] = 0
mapping[v] += 1
value_set = set()
for v in mapping:
if mapping[v] in value_set:
return False
value_set.add(mapping[v])
return True
我是廢物寫好長，但是過ㄌGP
Javascript
function uniqueOccurrences(arr) {
const map = new Map();
for (const num of arr) {
map.set(num, (map.get(num) || 0) + 1);
}
return map.size === new Set(map.values()).size;
}
昨天學會用map跟set真的是太好用惹！
Marsgoat Nov 30, 2022
TypeScript
function uniqueOccurrences(arr: number[]): boolean {
const map = new Map();
arr.forEach((key) => {
map.set(key, (map.get(key) ?? 0) + 1);
});
// map.size 跟 value 組成的 set.size 不一樣就代表有重複
return map.size === new Set(map.values()).size;
}
/*
* Map.prototype.set(key, value)：
* 設置鍵名 key 對應的鍵值為 value，然後回傳整個 Map 結構。
* 如果 key 已經有值，則鍵值會被更新，否則就新生成該鍵。
*
* Map.prototype.get(key)：
* 讀取 key 對應的鍵值，如果找不到 key，回傳 undefined。
*/
the 抄
sheep Nov 30, 2022
