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.

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

Solution1:

int majorityElement(vector<int> &num) {
        int len = num.size();
        for(int i = 0; i <= len / 2; i++){
            for(int j = i + 1; j < len - i; j++){
                if(num[j] < num[i]){
                    int temp = num[i]; 
                    num[i] = num[j];
                    num[j] = num[i];
                }
            }
        }
        return num[len/2];
    }

结果超时..参照https://leetcode.com/discuss/19151/solution-computation-space-problem-can-extended-situation解题思路，可得代码如下

Solution2:

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