【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;
}
};