[LeetCode]Majority Element

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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

 

最简单的方法就是排序，然后中间的那个数字就是我们需要的结果，两行搞定，时间复杂度就是排序的最优值O(nlogn)。

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        return nums[nums.size()/2];
    }
};

利用Hash Table可以实现O(n)的时间复杂度。

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        unordered_map<int,int> showed;
        for(int i=0;i<nums.size();i++)
        {
            if(showed.find(nums[i])==showed.end())
            {
                showed[nums[i]]=1;
            }
            else
            {
                showed[nums[i]]++;
            }
        }
        unordered_map<int,int>::iterator iter;
        for(iter=showed.begin();iter!=showed.end();iter++)
        {
            if(iter->second>nums.size()/2)
            {
                return iter->first;
            }
        }
    }
};

还有一种比较巧的方法，moore voting算法，时间复杂度也是O(n)。

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        int result;
        int count=0;
        for(int i=0;i<nums.size();i++)
        {
            if(count==0)
            {
                result=nums[i];
                count=1;
            }
            else
            {
                if(nums[i]==result) count++;
                else count--;
            }
        }
        return result;
    }
};

这是官网给的解法提示。