【LeetCode 229】Majority Element II

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space.

思路：

　　【LeetCode 169】Majority Element 的拓展，这回要求的是出现次数超过三分之一次的数字咯，动动我们的大脑思考下，这样的数最多会存在几个呢，当然是2个嘛。因此，接着上一题的方法做，只不过这回要投两个票啦，而且最后还得检查这两个投票结果是不是真的满足都超过三分之一，因为这一题题目什么都没有保证，所以答案可能有0个、1个、2个。

C++：

 class Solution {
 public:
     vector<int> majorityElement(vector<int>& nums) {
         vector<int> ret;

         int len = nums.size();
         if(len == )
             return ret;

         int m = , n = , cm = , cn = ;
         for(int i = ; i < len; i++)
         {
             int val = nums[i];
             if(m == val)
                 cm++;
             else if(n == val)
                 cn++;
             else if(cm == )
             {
                 m = val;
                 cm = ;
             }
             else if(cn == )
             {
                 n = val;
                 cn = ;
             }
             else
             {
                 cm--;
                 cn--;
             }
         }

         cm = cn = ;

         for(int i = ; i < len; i++)
         {
             if(nums[i] == m)
                 cm++;
             else if(nums[i] == n)
                 cn++;
         }

         if(cm *  > len)
             ret.push_back(m);
         if(cn *  > len)
             ret.push_back(n);

         return ret;
     }
 };

Python：

 class Solution:
     # @param {integer[]} nums
     # @return {integer[]}
     def majorityElement(self, nums):
         m, n, cm, cn = 0, 0, 0, 0
         ret = []

         for val in nums:
             if m == val:
                 cm = cm + 1
             elif n == val:
                 cn = cn + 1
             elif cm == 0:
                 m = val
                 cm = 1
             elif cn == 0:
                 n = val
                 cn = 1
             else:
                 cm = cm - 1
                 cn = cn - 1

         cm, cn = 0, 0

         for val in nums:
             if m == val:
                 cm = cm + 1
             elif n == val:
                 cn = cn + 1

         if cm * 3 > len(nums):
             ret.append(m)
         if cn * 3 > len(nums):
             ret.append(n)

         return ret