【LeetCode】137. Single Number II (3 solutions)

Single Number II

Given an array of integers, every element appears threetimes except for one. Find that single one.

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

 

解法一：开辟map记录次数

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        map<int, int> m;
        int n = nums.size();
        for(int i = 0; i < n; i ++)
        {
            if(m.find(nums[i]) == m.end())
                m[nums[i]] = 1;
            else
                m[nums[i]] ++;
        }
        for(map<int, int>::iterator it = m.begin(); it != m.end(); it ++)
        {
            if(it->second != 3)
                return it->first;
        }
    }
};

 

解法二：先排序，再遍历找出孤异元素

class Solution 
{
public:
    int singleNumber(vector<int>& nums) 
    {
        int n = nums.size();
        sort(nums.begin(), nums.end());
        //note that at least 4 elements in nums
        if(nums[0] != nums[1])
            return nums[0];
        if(nums[n-1] != nums[n-2])
            return nums[n-1];
        for(int i = 1; i < n-1; i ++)
        {
            if(nums[i] != nums[i-1] && nums[i] != nums[i+1])
                return nums[i];
        }
    }
};

 

解法三：位操作。不管非孤异元素重复多少次，这是通用做法。

对于右数第i位，如果孤异元素该位为0，则该位为1的元素总数为3的整数倍。

如果孤异元素该位为1，则该位为1的元素总数不为3的整数倍（也就是余1）。

换句话说，如果第i位为1的元素总数不为3的整数倍，则孤异数的第i位为1，否则为0.

（如果非孤异元素重复n次，则判断是否为n的整数倍）

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int ret = 0;
        int mask = 1;
        while(mask)
        {
            int countOne = 0;   //number of digit 1
            for(int i = 0; i < nums.size(); i ++)
            {
                if(nums[i] & mask)
                    countOne ++;
            }
            if(countOne % 3 == 1)
                ret |= mask;
            mask <<= 1;
        }
        return ret;
    }
};