Description#
You are given an integer array nums
. A number x
is lonely when it appears only once, and no adjacent numbers (i.e. x + 1
and x - 1)
appear in the array.
Return all lonely numbers in nums
. You may return the answer in any order.
Example 1:
Input: nums = [10,6,5,8]
Output: [10,8]
Explanation:
- 10 is a lonely number since it appears exactly once and 9 and 11 does not appear in nums.
- 8 is a lonely number since it appears exactly once and 7 and 9 does not appear in nums.
- 5 is not a lonely number since 6 appears in nums and vice versa.
Hence, the lonely numbers in nums are [10, 8].
Note that [8, 10] may also be returned.
Example 2:
Input: nums = [1,3,5,3]
Output: [1,5]
Explanation:
- 1 is a lonely number since it appears exactly once and 0 and 2 does not appear in nums.
- 5 is a lonely number since it appears exactly once and 4 and 6 does not appear in nums.
- 3 is not a lonely number since it appears twice.
Hence, the lonely numbers in nums are [1, 5].
Note that [5, 1] may also be returned.
Constraints:
1 <= nums.length <= 105
0 <= nums[i] <= 106
Solutions#
Solution 1#
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| class Solution:
def findLonely(self, nums: List[int]) -> List[int]:
counter = Counter(nums)
ans = []
for num, cnt in counter.items():
if cnt == 1 and counter[num - 1] == 0 and counter[num + 1] == 0:
ans.append(num)
return ans
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| class Solution {
public List<Integer> findLonely(int[] nums) {
Map<Integer, Integer> counter = new HashMap<>();
for (int num : nums) {
counter.put(num, counter.getOrDefault(num, 0) + 1);
}
List<Integer> ans = new ArrayList<>();
counter.forEach((k, v) -> {
if (v == 1 && !counter.containsKey(k - 1) && !counter.containsKey(k + 1)) {
ans.add(k);
}
});
return ans;
}
}
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| class Solution {
public:
vector<int> findLonely(vector<int>& nums) {
unordered_map<int, int> counter;
for (int num : nums) ++counter[num];
vector<int> ans;
for (auto& e : counter) {
int k = e.first, v = e.second;
if (v == 1 && !counter.count(k - 1) && !counter.count(k + 1)) ans.push_back(k);
}
return ans;
}
};
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| func findLonely(nums []int) []int {
counter := make(map[int]int)
for _, num := range nums {
counter[num]++
}
var ans []int
for k, v := range counter {
if v == 1 && counter[k-1] == 0 && counter[k+1] == 0 {
ans = append(ans, k)
}
}
return ans
}
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| function findLonely(nums: number[]): number[] {
let hashMap: Map<number, number> = new Map();
for (let num of nums) {
hashMap.set(num, (hashMap.get(num) || 0) + 1);
}
let ans: Array<number> = [];
for (let [num, count] of hashMap.entries()) {
if (count == 1 && !hashMap.get(num - 1) && !hashMap.get(num + 1)) {
ans.push(num);
}
}
return ans;
}
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