OJ 169 Majority Element

题目要求：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.
我的解法：

 1 int majorityElement(vector<int> &num) {
 2     int key = num.back();
 3     int k = num.size();
 4     k--;
 5     int i = k-1;
 6     while (i >= k/2)
 7     {
 8         if (num[i] != key)
 9         {
10             num[i] = num[k-1];
11             k--;
12             key = num[k-1];
13             k--;
14             i=k-1;
15             continue;
16         }
17         i--;
18     }
19     return key;
20 }

分析：首先要知道一个定理，如果一个序列中存在一个出现率大于1/2的元素，我们记为mE，那么如果去掉这个序列中的两个不相同的元素，新序列中一定存在一个出现率大于1/2的元素，依然是mE。
该算法最差情况需要在i指向第一个元素的时候，也就是算法的复杂度最差是O(n)，最好的情况需要n/2。

这个问题还可以直接考虑，通过map来计数，来获取所求的元素。代码如下：

 1 int majorityElement(vector<int> &num) {
 2     map<int, int> count;
 3     for (int i : num)
 4     {
 5         if (++count[i] >= num.size() / 2)
 6         {
 7             return i;
 8         }
 9     }
10 }

转载于:https://www.cnblogs.com/charleschiu/p/4338250.html

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