标签：
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
【LeetCode 229】Majority Element II
标签：
原文地址：http://www.cnblogs.com/tjuloading/p/4614424.html
