专题：数组中数字出现的次数

利用异或的特性

1. a ^ a =   0

2. a ^ 0 = a

3. a = a ^ b ^ b

题目一、数组中唯一只出现一次的数字

一次异或，得到最后的数字即为答案

int FindOnceNum(int arr[], int len)
{
	if (arr == NULL || len <= 0)
		return -1;
	int ret = 0;
	for (int i = 0; i<len; i++)
	{
		ret = ret^arr[i];
	}
	return ret;
}

题目二、数组中有2个只出现1次的数字

2,4,3,6,3,2,5,5

假设只出现1次的两个数字分别为a，b；则最后结果分别为c = a^b;

如题，异或结果为0010，则按照倒数第二位为1或者0将数组分为{2,3,6,3,2}和{4,5,5}

分别异或可得；

基于map的实现

class Solution {
public:
	void FindNumsAppearOnce(vector<int> data, int* num1, int *num2) {
		map<int, int> hashMap;
		for (int i = 0; i < data.size(); i++)
		{
			if (hashMap.count(data[i]))
				hashMap[data[i]]++;
			hashMap.insert(make_pair(data[i], 1));
		}
		auto piter = hashMap.begin();
		bool first = false;
		while (piter !=hashMap.end())
		{
			if (1 == piter->second && first == false)
			{
				*num1 = piter->first;
				first = true;
			}
			
			if (1 == piter->second && first == true)
			{
				*num2 = piter->first;
			}
			++piter;
		}
	}
};

基于异或运算的实现

class Solution {
public:
    void FindNumsAppearOnce(vector<int> data, int* num1, int *num2) {
        int len = data.size();
        int res = FindOnceNum(data, len);
        unsigned index = FindFirstBitIs1(res);
 
        *num1 = 0, *num2 = 0;
        for (int i = 0; i < len; i++)
        {
            if (!BitIs1(data[i], index))
                *num1 = *num1^data[i];
            else
                *num2 = *num2^data[i];
 
        }
    }
 
    unsigned FindFirstBitIs1(int num)  //右边第几位是1
    {
        unsigned index = 0;
        while (!(num & 1) && index <= 8 * sizeof(int))
        {
            num = num >> 1;
            ++index;
        }
        return index;
    }
 
    bool BitIs1(int num, int index)//右边数第index是不是1
    {
        num = num >> index;
        return (num & 1);
    }
    int FindOnceNum(vector<int> arr, int len)
    {
        if ( len <= 0)
            return -1;
        int ret = 0;
        for (int i = 0; i<len; i++)
        {
            ret = ret^arr[i];
        }
 
        return ret;
    }
};