Interview Question 39: Numbers with more than half of the occurrences in the array (C ++)

Title address: https://leetcode-cn.com/problems/shu-zu-zhong-chu-xian-ci-shu-chao-guo-yi-ban-de-shu-zi-lcof/

Title description

There is a number in the array that appears more than half the length of the array. Please find this number. You can assume that the array is non-empty, and there will always be a majority of elements in a given array.

Example questions

Example 1:

Input: [1, 2, 3, 2, 2, 2, 5, 4, 2]
Output: 2

Problem-solving ideas

Idea 1: Sort the array first, and then take the middle number, because the middle element must be more than half of the number after sorting.

Idea 2: The hash table counts the number of occurrences of each number, and returns the element when the number of occurrences exceeds half.

Idea 3: Use the voting method to conduct elections, each element can participate in the majority selection, if you encounter the same as yourself, you will add one vote, that is vote ++, if you encounter a different vote from yourself, that is vote--, and the final What comes down is what the title requires.

Idea 4: More than half of the numbers appear more often than all other numbers combined

Program source code

Idea 1

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

Idea 2

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        unordered_map<int, int> mp;
        for(int i = 0; i < nums.size(); i++)
        {
            mp[nums[i]]++;
        }
        for(int j = 0; j < nums.size(); j++)
        {
            if(mp[nums[j]] > nums.size() / 2) return nums[j];
        }
        return 0;
    }
};

Idea 3

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

Idea 4

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