• 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;
    		}
    };
    
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  • 原文地址:https://www.cnblogs.com/siweihz/p/12287120.html
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