544. Top k Largest Numbers【medium】

Given an integer array, find the top k largest numbers in it.

 

Example

Given [3,10,1000,-99,4,100] and k = 3.
Return [1000, 100, 10].

解法一：

 class Solution {
 public:
     /*
      * @param nums: an integer array
      * @param k: An integer
      * @return: the top k largest numbers in array
      */
     vector<int> topk(vector<int> nums, int k) {
         std::priority_queue<int> heap;
         for(int i = ; i < nums.size(); i++) {
             heap.push(nums[i]);
         }

         std::vector<int> result;
         for (int i = ; i < k; i++) {
             result.push_back(heap.top());
             heap.pop();
         }

         return result;
     }
 };

优先队列，类似于维护一个大小为k的最大堆/最小堆（STL中的priority_queue默认是最大堆），时间复杂度为O(n * logk)

解法二：

 class Solution {
     /*
      * @param nums an integer array
      * @param k an integer
      * @return the top k largest numbers in array
      */
     public int[] topk(int[] nums, int k) {
         int[] temp = new int[k];
         if(nums == null || nums.length == 0) {
             return temp;
         }
         quickSort(nums, 0, nums.length - 1, k);

         if(nums.length < k) {
             return nums;
         }

         for(int i = 0; i < k; i++) {
             temp[i] = nums[i];
         }

         return temp;
     }

     public void quickSort(int[] nums, int start, int end, int k) {
         if (start >= end) {
             return;
         }

         int left = start, right = end;
         int mid = left + (right - left) / 2;
         int pivot = nums[mid];

         while(left <= right) {
             while(left <= right && nums[left] > pivot) {
                 left++;
             }

             while(left <= right && nums[right] < pivot) {
                 right--;
             }

             if(left <= right) {
                 swap(nums, left, right);
                 left++;
                 right--;
             }
         }

         quickSort(nums, start, right, k);
         quickSort(nums, left, end, k);
     }

     private void swap(int[] nums, int left, int right) {
         int temp = nums[left];
         nums[left] = nums[right];
         nums[right] = temp;
     }
 };

快速排序，取出前k大的数，时间复杂度O(n*logn + k)