【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++：

 1 class Solution {
 2 public:
 3     vector<int> majorityElement(vector<int>& nums) {
 4         vector<int> ret;
 5
 6         int len = nums.size();
 7         if(len == 0)
 8             return ret;
 9
10         int m = 0, n = 0, cm = 0, cn = 0;
11         for(int i = 0; i < len; i++)
12         {
13             int val = nums[i];
14             if(m == val)
15                 cm++;
16             else if(n == val)
17                 cn++;
18             else if(cm == 0)
19             {
20                 m = val;
21                 cm = 1;
22             }
23             else if(cn == 0)
24             {
25                 n = val;
26                 cn = 1;
27             }
28             else
29             {
30                 cm--;
31                 cn--;
32             }
33         }
34
35         cm = cn = 0;
36
37         for(int i = 0; i < len; i++)
38         {
39             if(nums[i] == m)
40                 cm++;
41             else if(nums[i] == n)
42                 cn++;
43         }
44
45         if(cm * 3 > len)
46             ret.push_back(m);
47         if(cn * 3 > len)
48             ret.push_back(n);
49
50         return ret;
51     }
52 };

Python：

 1 class Solution:
 2     # @param {integer[]} nums
 3     # @return {integer[]}
 4     def majorityElement(self, nums):
 5         m, n, cm, cn = 0, 0, 0, 0
 6         ret = []
 7
 8         for val in nums:
 9             if m == val:
10                 cm = cm + 1
11             elif n == val:
12                 cn = cn + 1
13             elif cm == 0:
14                 m = val
15                 cm = 1
16             elif cn == 0:
17                 n = val
18                 cn = 1
19             else:
20                 cm = cm - 1
21                 cn = cn - 1
22
23         cm, cn = 0, 0
24
25         for val in nums:
26             if m == val:
27                 cm = cm + 1
28             elif n == val:
29                 cn = cn + 1
30
31         if cm * 3 > len(nums):
32             ret.append(m)
33         if cn * 3 > len(nums):
34             ret.append(n)
35
36         return ret