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Description
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
这道题给了一个整型数组，问每个数字出现的次数都否都不同。对于一道 Easy 的题目来说，没有太大的难度，就是用个 HashMap 来统计每个数字出现的次数，然后再用个 HashSet 来判断某个次数是否之前出现过了，若出现过了，则返回 false，否则最终返回 true 即可，参见代码如下：
class Solution {
public:
bool uniqueOccurrences(vector<int>& arr) {
unordered_map<int, int> numCnt;
unordered_set<int> st;
for (int num : arr) ++numCnt[num];
for (auto a : numCnt) {
if (st.count(a.second)) return false;
st.insert(a.second);
}
return true;
}
};
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