[C++]LeetCode: 50 Majority Element

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.

找出數組中出現次數超過數組長度一半的值。

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: 該算法要求目標數組存在majority元素（大於n/2），否則需要檢驗。 算法演示 here. 思路解析： 1. 初始化majorityIndex，並且維護其對應count； 2. 遍歷數組，如果下一個元素和當前候選元素相同，count加1，否則count減1； 3. 如果count為0時，則更改候選元素，並且重置count為1； 4. 返回A[majorityIndex] 原理：如果majority元素存在（majority元素個數大於n/2,個數超過數組長度一半），那麼無論它的各個元素位置是如何分布的，其count經過抵消和增加後，最後一定是大於等於1的。 如果不能保證majority存在，需要檢驗。 復雜度：O(N) Attention: 循環時從i = 1開始，從下一個元素開始，因為count已經置1. AC Code:

         class Solution {
         public:
             int majorityElement(vector &num) {
                 //the majority element 存在並且唯一
                 
                 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];
             }
         };

         
         檢驗： /* Function to check if the candidate occurs more than n/2 times */ boolisMajority(inta[], intsize, intcand) { inti, count = 0; for(i = 0; i < size; i++) if(a[i] == cand) count++; if(count > size/2) return1; else return0; }