刷题215. Kth Largest Element in an Array

一、题目说明

题目215. Kth Largest Element in an Array，在一个无序数组中找第k大的元素。难度是Medium！

二、我的解答

这个题目最直观的解答是，先对数组排序，然后直接返回：

class Solution{
	public:
		int findKthLargest(vector<int>& nums,int k){
			sort(nums.begin(),nums.end());
			
			return nums[nums.size()-k];
		}
};

性能如下：

Runtime: 8 ms, faster than 97.67% of C++ online submissions for Kth Largest Element in an Array.
Memory Usage: 9.2 MB, less than 93.94% of C++ online submissions for Kth Largest Element in an Array.

三、优化措施

用小根堆实现，无需多言：

class Solution{
	public:
		//queue first in first out with priority delete
		int findKthLargest(vector<int>& nums,int k){
			//升序队列 ，小根堆 
			priority_queue<int,vector<int>,greater<int> > q;
			
			for(auto it: nums){
				q.push(it);
				if(q.size()>k){
					cout<<q.top()<<"->"; 
					q.pop();
				}
			}
			
			return q.top();
		}
};

Runtime: 12 ms, faster than 80.01% of C++ online submissions for Kth Largest Element in an Array.
Memory Usage: 9.5 MB, less than 39.39% of C++ online submissions for Kth Largest Element in an Array.

上面2个方法都不是最好的办法：方法1胜在简单易实现，方法2直观，方法3利用快速排序的思想：

class Solution{
	public:
		//利用快速排序的思想，不断将集合划分为左右两部分，
		//如果划分的位置pivot>k-1，则第k大的数在左边 
		//如果划分的位置pivot<k-1，则第k大的数在右边 
		int findKthLargest(vector<int>& nums,int k){
			int low = 0,high = nums.size()-1,mid = 0;
			while(low<=high){
				mid = partation(nums,low,high);
				if(mid == k-1){
					return nums[mid];
				}else if(mid<k-1){
					low = mid + 1;
				}else{
					high = mid -1;
				}
			}
			return -1;
		}
		int partation(vector<int>& nums,int low,int high){
			int left = low+1;
			int right = high;
			swap(nums[low],nums[(low+high)/2]);
			int bound = nums[low];
			while(left<=right){
				while(left<high && nums[left] >= bound) left++;
				while(nums[right]<bound) right--;
				if(left<right){
					swap(nums[left++],nums[right--]);
				}else{
					break;
				}
			}
			swap(nums[low],nums[right]);
			return right;
		}
};