[LeetCode] 136. Single Number 单独数

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

由于要求时间复杂度O(n)，空间复杂度O(1)，所以不能用排序法，也不能使用map。

解法1：用两倍所有非重复元素和减去原数组。

解法2：位操作Bit Operation，使用二进制数位操作中的异或，同为0，异为1。主要考察位操作。

异或运算{displaystyle Aoplus B}的真值表如下： F表示false，T表示true 

A	B	⊕
F	F	F
F	T	T
T	F	T
T	T	F

Java:

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

Python:

def singleNumber(self, nums):
    """
    :type nums: List[int]
    :rtype: int
    """
    return 2 * sum(set(nums)) - sum(nums)　　

Python:

class Solution(object):
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        res = 0
        for num in nums:
            res ^= num
            
        return res　　

Python:

import operator
from functools import reduce

class Solution:
    """
    :type nums: List[int]
    :rtype: int
    """
    def singleNumber(self, A):
        return reduce(operator.xor, A)　　

C++:

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int res = 0;
        for (auto num : nums) res ^= num;
        return res;
    }
};

类似题目：

[LeetCode] 137. Single Number II 单独数 II

[LeetCode] 260. Single Number III 单独数 III

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