leetcode--Majority Element

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.

解题思路:先给数组排序,然后找到符合间隔的。

c++版本“

#include <iostream>
#include<vector>
#include<algorithm>
using namespace std;

int majorityElement(vector<int>& nums)
{
    int n=nums.size()%2==0?nums.size()/2:(nums.size()/2+1);
    if (nums.size()==1) return nums[0];
    sort(nums.begin(),nums.end());
    for (int i=0; i<n; i++)
    {
        if(nums[i]==nums[i+nums.size()/2])
            return nums[i];
    }
    return 0;
}

int main()
{
    vector <int> s;
    s.push_back(1);
    s.push_back(1);
    s.push_back(1);
    s.push_back(2);
    s.push_back(3);
    int sum=majorityElement(s);
    cout<<sum<<endl;
}

java版本:

public class Solution {
    public int majorityElement(int[] nums) {
        if(nums.length==1) return nums[0];
        int n=nums.length%2==0?nums.length/2:(nums.length/2+1);
        Arrays.sort(nums);
        for (int i=0;i<n;i++){
          if(nums[i]==nums[i+nums.length/2])
           return nums[i];
    }
        return 0;
    } 
}

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