[LeetCode] 169. Majority Element
Description
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.
Example 1:
Input: [3,2,3]
Output: 3
Example 2:
Input: [2,2,1,1,1,2,2]
Output: 2
Analyse
找出一个长为\(n\)的数组中出现次数大于⌊ n/2 ⌋的数
\(n\)不为0
majority element一定存在
第一种方法，majority element的出现次数大于一半，先排序，然后直接取中间的元素，时间复杂度主要是sort的，平均时间复杂度为\(O(Nlog(N))\)
The C standard doesn’t talk about its complexity of qsort. The new C++11 standard requires that the complexity of sort to be O(Nlog(N)) in the worst case. Previous versions of C++ such as C++03 allow possible worst case scenario of O(N^2). Only average complexity was required to be O(N log N).
int majorityElement(vector<int>& nums)
{
sort(nums.begin(), nums.end());
return nums[nums.size() / 2];
}
结果被大佬们虐成渣了
Runtime: 28 ms, faster than 28.19% of C++ online submissions for Majority Element.
Memory Usage: 11.1 MB, less than 80.66% of C++ online submissions for Majority Element.
第二种方法，计数法，找个map计算元素出现次数，输出次数大于一半的元素，时间复杂度\(O(n)\)
int majorityElement(vector<int>& nums)
{
unordered_map<int, int> map;
for(int a : nums)
{
if (++map[a] > (nums.size() / 2))
{
return a;
}
}
return -1;
}
比前面那种方法快了一点，但在leetcode上还是不够快，
Runtime: 24 ms, faster than 48.33% of C++ online submissions for Majority Element.
Memory Usage: 11 MB, less than 96.80% of C++ online submissions for Majority Element.
加上一些优化代码后，在leetcode上已经是最快了
static int x=[](){
std::ios::sync_with_stdio(false);
cin.tie(NULL);
return 0;
}();
class Solution {
public:
int majorityElement(vector<int>& nums)
{
unordered_map<int, int> map;
for(int a : nums)
{
if (++map[a] > (nums.size() / 2))
{
return a;
}
}
return -1;
}
};
Runtime: 4 ms, faster than 100.00% of C++ online submissions for Majority Element.
Memory Usage: 11.1 MB, less than 74.76% of C++ online submissions for Majority Element.
第三种方法，来自leetcode，由于majority element的个数大于一半，对数组中不一样的元素两两删除，剩下的就是majority element
直接对vector删除元素很耗时，使用stack来模拟这个删除
int majorityElement(vector<int>& nums)
{
stack<int> tony;
for(int a : nums)
{
if (tony.empty() || a == tony.top())
{
tony.push(a);
}
else
{
tony.pop();
}
}
return tony.top();
}
Runtime: 8 ms, faster than 99.91% of C++ online submissions for Majority Element.
Memory Usage: 11.7 MB, less than 5.43% of C++ online submissions for Majority Element.
