[Algorithm] 5. Kth Largest Element

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

     /**
      * @param n: An integer
      * @param nums: An array
      * @return: the Kth largest element
      */
     int kthLargestElement(int n, vector<int> &nums) {
         // find out the kth largest number in an array.
         int temp=;
         for(int i=; i<=n; i++)
         {
             for(int j=; j<nums.size()-i; j++)
             {
                 // Swap the larger number to the end.
                 if(nums[j] >= nums[j+])
                 {
                     temp = nums[j];
                     nums[j] = nums[j+];
                     nums[j+] = temp;
                 }
             }
         }

         return nums[nums.size()-n];
     }

Better Solution

     /**
      * @param n: An integer
      * @param nums: An array
      * @return: the Kth largest element
      */

     int quick_sort(vector<int> &nums, int l, int r)
     {
         int key=nums[l];
         while(l<r)
         {
             while( (nums[r] >= key) && (l<r) )  r--;
             swap(nums[l], nums[r]);

             while( (nums[l] <= key) && (l<r) )  l++;
             swap(nums[l], nums[r]);
         }

         return l;
     }

     int kthLargestElement(int n, vector<int> &nums) {
         // Using qsort method to find the kth largest number.
         if(n>nums.size() || n== || nums.size()==)   return -;

         int left=, right=nums.size()-, k=nums.size()-n, pivot=;

         while(true){
             pivot = quick_sort(nums, left, right);
             if(pivot==k){
                 return nums[k];
             }else if(pivot<k){
                 left=pivot+;
                 right=nums.size()-;
             }else{
                 left=;
                 right=pivot-;
             }
         }
     }

Best Solution

     /**
      * @param n: An integer
      * @param nums: An array
      * @return: the Kth largest element
      */
     int kthLargestElement(int n, vector<int> &nums) {
         // Using priority queue to solve it.
         if(n>nums.size() || n== || nums.size()==)   return -;

         priority_queue<int, vector<int>, greater<int>> pri_queue;
         for(int i=; i<n; i++)
         {
             pri_queue.push(nums[i]);
         }
         for(int i=n; i<nums.size(); i++)
         {
             if( nums[i] > pri_queue.top() )
             {
                 pri_queue.pop();
                 pri_queue.push(nums[i]);
             }
         }

         return pri_queue.top();
     }

Tips

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