LeetCode【169. 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.

 

思路.1

排序，选择第n/2个数，调用STL的sort,所以时间复杂度是O(nlogn)

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

　　

思路2

设置一个计数，当count为0时设置majority的值，如果下一个值与majority相同则count+1,如果不同，则-1，当count==0时，对majority重新赋值，因为majority的个数肯定大于n/2所以最后>0的count的肯定是majority，这样，就只需要遍历一遍就可以求出majority，时间复杂度为O(n)

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