[C++]LeetCode: 50 Majority Element

题目：

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.

找出数组中出现次数超过数组长度一半的值。

Solution:

1. Runtime: O(n2)— Brute force solution: Check each element if it is the majority element.
2. Runtime: O(n), Space: O(n)— Hash table: Maintain a hash table of the counts of each element, then find the most common one.

1. Runtime: O(n log n)— Divide and conquer: Divide the array into two halves, then find the majority element A in the first half and the majority element B in the second half. The global majority element must either be A or B. If A == B, then it automatically becomes the global majority element. If not, then both A and B are the candidates for the majority element, and it is suffice to check the count of occurrences for at most two candidates. The runtime complexity, T(n) = T(n/2) + 2n = O(n logn).
2. Runtime: O(n) —Moore voting algorithm: We maintain a current candidate and a counter initialized to 0. As we iterate the array, we look at the current element x:
   
   1. If the counter is 0, we set the current candidate to x and the counter to 1.
   2. If the counter is not 0, we increment or decrement the counter based on whether x is the current candidate.
   After one pass, the current candidate is the majority element. Runtime complexity = O(n)

Moore Voting Algorithm:

该算法要求目标数组存在majority元素（大于n/2），否则需要检验。

算法演示 here. 

思路解析：

1. 初始化majorityIndex，并且维护其对应count；

2. 遍历数组，如果下一个元素和当前候选元素相同，count加1，否则count减1；

3. 如果count为0时，则更改候选元素，并且重置count为1；

4. 返回A[majorityIndex]

原理：如果majority元素存在（majority元素个数大于n/2,个数超过数组长度一半），那么无论它的各个元素位置是如何分布的，其count经过抵消和增加后，最后一定是大于等于1的。

如果不能保证majority存在，需要检验。

复杂度：O(N)

Attention: 循环时从i = 1开始，从下一个元素开始，因为count已经置1.

AC Code:

class Solution {
public:
    int majorityElement(vector<int> &num) {
        //the majority element 存在并且唯一
        
        int majorityIndex = 0;
        for(int cnt = 1, i = 1; i < num.size(); i++)
        {
            num[majorityIndex] == num[i] ? cnt++ : cnt--;
            if(cnt == 0)
            {
                cnt = 1;
                majorityIndex = i;
            }
        }
        
        return num[majorityIndex];
    }
};

检验：

/* Function to check if the candidate occurs more than n/2 times */
bool isMajority( int a[], int size, int cand)
{
    int i, count = 0;
    for (i = 0; i < size; i++)
      if (a[i] == cand)
         count++;
    if (count > size/2)
       return 1;
    else
       return 0;
}