LeetCode 솔루션 분류
[Easy - wk6 - Q2] 136. Single Number
본문
Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
Example 1:
Input: nums = [2,2,1] Output: 1
Example 2:
Input: nums = [4,1,2,1,2] Output: 4
Example 3:
Input: nums = [1] Output: 1
Constraints:
1 <= nums.length <= 3 * 104-3 * 104 <= nums[i] <= 3 * 104- Each element in the array appears twice except for one element which appears only once.
태그
#leetcode, #easy, #문제풀이, #amazon, #adobe, #microsoft, #bloomberg, #apple, #facebook, #google, #uber, #atlassian, #tcs, #array, #bit manipulation
관련자료
댓글 3
namsoo님의 댓글
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class Solution:
def singleNumber(self, nums: List[int]) -> int:
nhash = defaultdict(int)
for n in nums:
nhash[n] += 1
for k, v in nhash.items():
if v == 1:
return k
return None
def singleNumber(self, nums: List[int]) -> int:
nhash = defaultdict(int)
for n in nums:
nhash[n] += 1
for k, v in nhash.items():
if v == 1:
return k
return None
학부유학생님의 댓글
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Runtime: 141 ms, faster than 85.70% of Python3 online submissions for Single Number.
Memory Usage: 16.7 MB, less than 84.62% of Python3 online submissions for Single Number.
Memory Usage: 16.7 MB, less than 84.62% of Python3 online submissions for Single Number.
class Solution:
def singleNumber(self, nums: List[int]) -> int:
res = 0
for num in nums: res ^= num
return res
# return 2*sum(set(nums)) - sum(nums)Coffee님의 댓글
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- 작성일
class Solution {
public int singleNumber(int[] nums) {
int result = 0;
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int i=0; i<nums.length; i++){
int num = map.getOrDefault(nums[i], 0);
map.put(nums[i], num+1);
}
for(Map.Entry<Integer, Integer> entry : map.entrySet()){
if(entry.getValue() == 1){
result = entry.getKey();
}
}
return result;
}
}
