【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