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.

 1 class Solution {
 2 public:
 3     int majorityElement(vector<int> &num) {
 4         int len = num.size(), result = 0 ,rec = 0;
 5         unordered_map<int, int> list;
 6         for (int i = 0; i < len; i++) {
 7             if (list[num[i]]) list[num[i]]++;
 8             else list[num[i]] = 1;
 9             if (rec < list[num[i]]) {
10                 rec = list[num[i]];
11                 result = num[i];
12             }
13         }
14         return result;
15     }
16 };

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;

 1 class Solution {
 2 public:
 3     int majorityElement(vector<int> &num) {
 4         sort(num.begin(), num.end());
 5         int len = num.size(), rec = 1, current = num[0];
 6         for (int i = 1; i < len; i++) {
 7            if (current != num[i]) {
 8                if (rec > len/2) {
 9                    return num[i-1];
10                }
11                current = num[i];
12                rec = 1;
13            } else {
14                rec++;
15            }
16         }
17         return num[len-1];
18     }
19 };