Description
You are given an integer array nums. The unique elements of an array are the elements that appear exactly once in the array.
Return the sum of all the unique elements of nums.
Example 1:
Input: nums = [1,2,3,2]
Output: 4
Explanation: The unique elements are [1,3], and the sum is 4.
Example 2:
Input: nums = [1,1,1,1,1]
Output: 0
Explanation: There are no unique elements, and the sum is 0.
Example 3:
Input: nums = [1,2,3,4,5]
Output: 15
Explanation: The unique elements are [1,2,3,4,5], and the sum is 15.
Constraints:
1 <= nums.length <= 1001 <= nums[i] <= 100
Solutions
Solution 1
PythonJavaC++GoTypeScriptRustPHP
class Solution:
def sumOfUnique(self, nums: List[int]) -> int:
cnt = Counter(nums)
return sum(x for x, v in cnt.items() if v == 1)(code-box)
class Solution {
public int sumOfUnique(int[] nums) {
int[] cnt = new int[101];
for (int x : nums) {
++cnt[x];
}
int ans = 0;
for (int x = 0; x < 101; ++x) {
if (cnt[x] == 1) {
ans += x;
}
}
return ans;
}
}(code-box)
class Solution {
public:
int sumOfUnique(vector<int>& nums) {
int cnt[101]{};
for (int& x : nums) {
++cnt[x];
}
int ans = 0;
for (int x = 0; x < 101; ++x) {
if (cnt[x] == 1) {
ans += x;
}
}
return ans;
}
};(code-box)
func sumOfUnique(nums []int) (ans int) {
cnt := [101]int{}
for _, x := range nums {
cnt[x]++
}
for x := 0; x < 101; x++ {
if cnt[x] == 1 {
ans += x
}
}
return
}(code-box)
function sumOfUnique(nums: number[]): number {
const cnt = new Array(101).fill(0);
for (const x of nums) {
++cnt[x];
}
let ans = 0;
for (let x = 0; x < 101; ++x) {
if (cnt[x] == 1) {
ans += x;
}
}
return ans;
}(code-box)
impl Solution {
pub fn sum_of_unique(nums: Vec<i32>) -> i32 {
let mut cnt = [0; 101];
for x in nums {
cnt[x as usize] += 1;
}
let mut ans = 0;
for x in 1..101 {
if cnt[x] == 1 {
ans += x;
}
}
ans as i32
}
}(code-box)
class Solution {
/**
* @param Integer[] $nums
* @return Integer
*/
function sumOfUnique($nums) {
$sum = 0;
for ($i = 0; $i < count($nums); $i++) {
$hashtable[$nums[$i]] += 1;
if ($hashtable[$nums[$i]] == 1) {
$sum += $nums[$i];
}
if ($hashtable[$nums[$i]] == 2) {
$sum -= $nums[$i];
}
}
return $sum;
}
}(code-box)
Solution 2
JavaC++GoTypeScriptRust
class Solution {
public int sumOfUnique(int[] nums) {
int ans = 0;
int[] cnt = new int[101];
for (int x : nums) {
if (++cnt[x] == 1) {
ans += x;
} else if (cnt[x] == 2) {
ans -= x;
}
}
return ans;
}
}(code-box)
class Solution {
public:
int sumOfUnique(vector<int>& nums) {
int ans = 0;
int cnt[101]{};
for (int& x : nums) {
if (++cnt[x] == 1) {
ans += x;
} else if (cnt[x] == 2) {
ans -= x;
}
}
return ans;
}
};(code-box)
func sumOfUnique(nums []int) (ans int) {
cnt := [101]int{}
for _, x := range nums {
cnt[x]++
if cnt[x] == 1 {
ans += x
} else if cnt[x] == 2 {
ans -= x
}
}
return
}(code-box)
function sumOfUnique(nums: number[]): number {
let ans = 0;
const cnt = new Array(101).fill(0);
for (const x of nums) {
if (++cnt[x] === 1) {
ans += x;
} else if (cnt[x] === 2) {
ans -= x;
}
}
return ans;
}(code-box)
use std::collections::HashMap;
impl Solution {
pub fn sum_of_unique(nums: Vec<i32>) -> i32 {
let mut res = 0;
let mut map = HashMap::new();
for num in nums {
if map.contains_key(&num) {
if *map.get(&num).unwrap() {
map.insert(num, false);
res -= num;
}
} else {
map.insert(num, true);
res += num;
}
}
res
}
}(code-box)
