【LeetCode】 Majority Element 系列

Majority Element 系列

169. Majority Element

介绍

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋times.

You may assume that the array is non-empty and the majority element always exist in the array.

解答

摩尔投票算法

class Solution {
public:
    //摩尔投票算法
    int majorityElement(vector<int> & nums)
    {
        int count = 1, major = nums[0], n = nums.size();
        for(int i = 1; i < n; ++i)
        {
            if(!count)  major = nums[i],count = 1;
            else if(nums[i] == major)    ++count;
            else --count;
        }
        return major;
    }
};

位操作

class Solution {
public:
    //Bit Manipulation
    int majorityElement(vector<int> & nums)
    {
        int bitCount = 0,majority = 0, n = nums.size();
        int mask = 1;
        while(mask)
        {
            int bitCount = 0;
            for(int j = 0; j < n; ++j)
            {
                if(nums[j] & mask)  ++bitCount;
                if(bitCount > n/2)  
                {
                    majority |= mask;
                    break;
                }
            }
            cout << mask << endl;
            mask <<= 1;
        }
        return majority;
    }  
};

哈希表

class Solution {
public:
   //哈希表
    int majorityElement(vector<int>& nums) {
        unordered_map<int,int> help;
        for(auto val : nums)
            ++help[val];
        for(auto val: help)
            if(val.second > nums.size()/2)
                return val.first;
    }
};

#### 分治法

```cpp

class Solution {
public:
    //Divide and Conquer 分治法
    int majorityElement(vector<int>& nums)
    {
        return getMajorityElement(nums,0,nums.size()-1);
    }
private:
    int getMajorityElement(const vector<int>& nums,int left,int right)
    {
        if(left == right)   return nums[left];
        int mid = (left+right)/2;
        int leftMajority = getMajorityElement(nums,left,mid);
        int rightMajority = getMajorityElement(nums,mid+1,right);
        if(leftMajority == rightMajority)   return leftMajority;
        return count(nums.begin()+left,nums.begin()+mid+1,leftMajority) > count(nums.begin()+mid+1,nums.begin()+right+1,rightMajority) ? leftMajority : rightMajority;
    }
};




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229. Majority Element II

介绍

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space.

解答

摩尔投票算法

class Solution {
public:
    // Boyer-Moore Majority Vote algorithm 摩尔投票算法
    vector<int> majorityElement(vector<int>& nums) {
        int count1 = 0,count2 = 0;
        int majority1 = 0, majority2 = 0,n =  nums.size();
        if(n == 0)  return vector<int>();
        for(int i = 0; i < n; ++i)
        {
            if(nums[i] == majority1)
                ++count1;
            else if (nums[i] == majority2)
                ++count2;
            else if(count1 == 0)
            {
                majority1 = nums[i];
                count1 = 1;
            }else if(count2 == 0)
            {
                majority2 = nums[i];
                count2 = 1;
            }
            else 
                --count1,--count2;
        }
        vector<int>  res;
        count1 = count2 = 0;
        for(auto val : nums)
        {
            if(val == majority1)    ++count1;
            if(val == majority2)    ++count2;
        }
        if(count1 > n/3)    res.push_back(majority1);
        if(majority1 != majority2)
        {
            if(count2 > n/3)    res.push_back(majority2);
        }
        return res;   
    }
};