【LeetCode】169

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.

 Solution 1: 使用map计数

class Solution {
public:
    int majorityElement(vector<int>& nums) {　　　　//runtime:32ms
        map<int, int> nums_count;
        vector<int>::iterator iter=nums.begin();
        
        while(iter!=nums.end()){
            ++nums_count[*iter];
            iter++;
        }
        
        int n=nums.size();
        
        map<int, int>::iterator ite=nums_count.begin();
        //int max=ite->second;
        while(ite!=nums_count.end()){
            if(ite->second>n/2){
                return ite->first;
            }
            ite++;
        }
    }
};

Solution 2: 每找出两个不同的element就成对删除，最后可能剩两个元素或一个元素，必定都是所求

class Solution {
public:
    int majorityElement(vector<int>& nums) {    //runtime: 20ms
        int count=0;
        int ret;
        for(int i=0;i<nums.size();i++){
            if(count==0){
                ret=nums[i];
                count++;
            }else{
                if(nums[i]==ret)
                    count++;
                else
                    count--;
            }
        }
        return ret;
    }
};

扩展到⌊ n/k ⌋的情况，每k个不同的element进行成对删除