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.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
直接上代码吧
1 class Solution { 2 public: 3 int majorityElement(vector<int>& nums) 4 { 5 int n = nums.size(); 6 sort(nums.begin(), nums.end()); 7 int cnt = 0; 8 int res = 0; 9 for (int i = 0; i < n; ++i) 10 { 11 int s = count(nums.begin(), nums.end(), nums[i]); 12 if (cnt < s) 13 { 14 cnt = s; 15 res = nums[i]; 16 } 17 i += s; 18 } 19 return res; 20 } 21 };
意思应该比较明了，就是不是很快速，也不简洁，看看其他大神的解答吧
这个哈希表应该能想到的呀~~~刷题比较少没有感觉~~~
class Solution { public: int majorityElement(vector<int>& nums) { unordered_map<int, int> counts; int n = nums.size(); for (int i = 0; i < n; i++) if (++counts[nums[i]] > n / 2) return nums[i]; } };
用到了nth_element函数，既然是多于一半的数自然中间.
class Solution { public: int majorityElement(vector<int>& nums) { nth_element(nums.begin(), nums.begin() + nums.size() / 2, nums.end()); return nums[nums.size() / 2]; } };
既然用了上面这个函数，那么我为什么不直接sort就行了=_+, 真的太笨了。
1 class Solution { 2 public: 3 int majorityElement(vector<int>& nums) 4 { 5 int n = nums.size(); 6 7 sort(nums.begin(), nums.end()); 8 9 return nums[n/2]; 10 } 11 };
先记这些吧~~~
