136. Single Number

Given a non-empty array of integers, every element appears twice except for one. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Example 1:

Input: [2,2,1]
Output: 1

Example 2:

Input: [4,1,2,1,2]
Output: 4

Approach #1: C++.

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        unordered_map<int, int> mp;
        for (int i = 0; i < nums.size(); ++i) 
            mp[nums[i]]++;
        for (int i = 0; i < nums.size(); ++i)
            if (mp[nums[i]] == 1) return nums[i];
    }
};

　　

Approach #2: Java.

class Solution {
    public int singleNumber(int[] nums) {
        int result = 0;
        for (int i : nums)
            result ^= i;
        return result;
    }
}

Note: ^ return 0 if two elements is same.

 

Approach #3: Python.

class Solution(object):
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        dic = {}
        for num in nums:
            dic[num] = dic.get(num, 0) + 1
        for key, val in dic.items():
            if val == 1:
                return key

The dictionary get method:

Description

The method get() returns a value for the given key. If key is not available then returns default value None.

Syntax

Following is the syntax for get() method −

dict.get(key, default = None)

Parameters

• key − This is the Key to be searched in the dictionary.

• default − This is the Value to be returned in case key does not exist.

Return Value

This method return a value for the given key. If key is not available, then returns default value None.

 

Time Submitted	Status	Runtime	Language
a few seconds ago	Accepted	36 ms	python
2 minutes ago	Accepted	1 ms	java
5 minutes ago	Accepted	12 ms	cpp