LeetCode – Refresh – Majority Element

1. Use memory :

In the question, it already stated that there must be a majority element. So discard empty situation.

Also it needs the number, not the index.

class Solution {
public:
    int majorityElement(vector<int> &num) {
        int len = num.size(), result = 0 ,rec = 0;
        unordered_map<int, int> list;
        for (int i = 0; i < len; i++) {
            if (list[num[i]]) list[num[i]]++;
            else list[num[i]] = 1;
            if (rec < list[num[i]]) {
                rec = list[num[i]];
                result = num[i];
            }
        }
        return result;
    }
};

 

2. With out memory:

Notes:

1. initialize the record after sorting. Because the ordering has been changed.

2. As question mentioned, majority over (n/2) times. Just use len/2 as the condition.

3. Return value is num[i-1] not num[i]. Because the num[i] is not the current value. Also we can put return current;

class Solution {
public:
    int majorityElement(vector<int> &num) {
        sort(num.begin(), num.end());
        int len = num.size(), rec = 1, current = num[0];
        for (int i = 1; i < len; i++) {
           if (current != num[i]) {
               if (rec > len/2) {
                   return num[i-1];
               }
               current = num[i];
               rec = 1;
           } else {
               rec++;
           }
        }
        return num[len-1];
    }
};