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[Easy - wk6 - Q2] 136. Single Number

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작성일 2022.06.07 01:07

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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.

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익명
작성일 2022.06.08 20:34

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

익명
작성일 2022.06.10 15:54

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.

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)

익명
작성일 2022.06.10 18:02

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;
    }
}