LeetCode Review on Interview 2018 - LRI

Chapter 1 - 1

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

Method 1 :

using hashset to cut the search time.

java code:

public static int singleNumber(int[] nums){
    if(nums.length==0) throw new IllegalArgumentException();
    HashSet<Integer> set = new HashSet<>();
    for(int i=0;i<nums.length;i++){
        if(set.contains(nums[i])) set.remove(nums[i]);
        else set.add(nums[i]);
    }
    if(set.size()>1) throw new IllegalArgumentException();
    else{
        Iterator<Integer> i = set.iterator();
        return i.next();
    }
}

Method 2:

XOR:

    如果我们对 0 和二进制位做 XOR 运算，得到的仍然是这个二进制位
    a⊕0=a 
如果我们对相同的二进制位做 XOR 运算，返回的结果是 0
    a⊕a=0
XOR 满足交换律和结合律
    a⊕b⊕a=(a⊕a)⊕b=0⊕b=b

所以我们只需要将所有的数进行 XOR 操作，得到那个唯一的数字。

java code:

public static int singleNumber(int[] nums){
    int a = 0;
    for(int i = 0;i<nums.length;i++){
        a ^= nums[i]; // ^ is for xor
    }
    return a;
}