标签:put nat code hash arrays inpu nts ret set
Given an array of integers arr
, write a function that returns true
if and only if the number of occurrences of each value in the array is unique.
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
class Solution { public boolean uniqueOccurrences(int[] arr) { Arrays.sort(arr); HashMap<Integer, Integer> map = new HashMap(); for(int i: arr){ map.put(i, 1 + map.getOrDefault(i, 0)); } ArrayList<Integer> list = new ArrayList(map.values()); Collections.sort(list); for(int i = 1; i < list.size(); i++){ if(list.get(i) == list.get(i-1)) return false; } return true; } }
绕远路还是我会绕。
HashMap的values() 方法返回一个values的set,形式为collections,所以可以转化为list
1207. Unique Number of Occurrences
标签:put nat code hash arrays inpu nts ret set
原文地址:https://www.cnblogs.com/wentiliangkaihua/p/11624005.html