leetcode 169. Majority Element 摩尔投票法-蒲公英云

leetcode 169. Majority Element 摩尔投票法

野性酷女 2022-06-08 07:28 261阅读 0赞

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.

就是使用HashMap直接计数。

代码如下：

import java.util.HashMap;
import java.util.Map;
public class Solution {
    public int majorityElement(int[] nums) 
    {
        Map<Integer, Integer> map=new HashMap<Integer, Integer>();
        for(int i=0;i<nums.length;i++)
            map.put(nums[i], map.getOrDefault(nums[i], 0)+1);
        for (Integer key : map.keySet()) 
        {  
            if(map.get(key) >= (nums.length+1)/2)
                return key;
        } 
        return 0;
    }
}

下面是C++的做法，就是使用map做一次遍历操作

代码如下：

#include <iostream>
#include <vector>
#include <string>
#include <map>
using namespace std;
class Solution
{
public:
    int majorityElement(vector<int>& nums)
    {
        map<int, int> mmp;
        for (int a : nums)
        {
            if (mmp.find(a) == mmp.end())
                mmp[a] = 1;
            else
                mmp[a] += 1;
        }
        for (map<int, int>::iterator i = mmp.begin(); i != mmp.end(); i++)
        {
            if (i->second >= (nums.size() + 1) / 2)
                return i->first;
        }
        return 0;
    }
};

其实这道题考察的是摩尔投票法，代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>
#include <iomanip>
#include <cstdlib>
#include <ctime>
using namespace std;
class Solution
{
public:
    int majorityElement(vector<int>& nums)
    {
        int count = 0, num = INT_MIN;
        for (int i : nums)
        {
            if (count == 0)
            {
                count = 1;
                num = i;
            }
            else if (i == num)
                count++;
            else
                count--;
        }
        return num;
    }
};