Description

Find K-th largest element in an array.

Example

In array [9,3,2,4,8], the 3rd largest element is 4.

In array [1,2,3,4,5], the 1st largest element is 5, 2nd largest element is 4, 3rd largest element is 3 and etc.

Challenge

O(n) time, O(1) extra memory.

Notice

You can swap elements in the array

Answer

 1     /**
 2      * @param n: An integer
 3      * @param nums: An array
 4      * @return: the Kth largest element
 5      */
 6     int kthLargestElement(int
 
 n, vector<int> &nums) {
 7         // find out the kth largest number in an array.
 8         int temp=0;
 9         for(int i=1; i<=n; i++)
10         {
11             for(int j=0; j<nums.size()-i; j++)
12             {
13                 // Swap the larger number to the end.
14                 if
 
(nums[j] >= nums[j+1])
15                 {
16                     temp = nums[j];
17                     nums[j] = nums[j+1];
18                     nums[j+1] = temp;
19                 }
20             }
21         }
22         
23         return nums[nums.size()-n];
24     }

Better Solution

 1     /**
 2      * @param n: An integer
 3      * @param nums: An array
 4      * @return: the Kth largest element
 5      */
 6      
 7     int quick_sort(vector<int> &nums, int l, int r)
 8     {
 9         int key=nums[l];
10         while(l<r)
11         {
12             while( (nums[r] >= key) && (l<r) )  r--;
13             swap(nums[l], nums[r]);
14             
15             while( (nums[l] <= key) && (l<r) )  l++;
16             swap(nums[l], nums[r]);
17         }
18         
19         return l;
20     }
21     
22     int kthLargestElement(int n, vector<int> &nums) {
23         // Using qsort method to find the kth largest number.
24         if(n>nums.size() || n==0 || nums.size()==0)   return -1;
25         
26         int left=0, right=nums.size()-1, k=nums.size()-n, pivot=0;
27         
28         while(true){
29             pivot = quick_sort(nums, left, right);
30             if(pivot==k){
31                 return nums[k];
32             }else if(pivot<k){
33                 left=pivot+1;
34                 right=nums.size()-1;
35             }else{
36                 left=0;
37                 right=pivot-1;
38             }
39         }
40     }

Best Solution

 1     /**
 2      * @param n: An integer
 3      * @param nums: An array
 4      * @return: the Kth largest element
 5      */
 6     int kthLargestElement(int n, vector<int> &nums) {
 7         // Using priority queue to solve it.
 8         if(n>nums.size() || n==0 || nums.size()==0)   return -1;
 9         
10         priority_queue<int, vector<int>, greater<int>> pri_queue;
11         for(int i=0; i<n; i++)
12         {
13             pri_queue.push(nums[i]);
14         }
15         for(int i=n; i<nums.size(); i++)
16         {
17             if( nums[i] > pri_queue.top() )
18             {
19                 pri_queue.pop();
20                 pri_queue.push(nums[i]);
21             }
22         }
23         
24         return pri_queue.top();
25     }

Tips

It is a good way to use STL container(priority queue) to solve the problem.