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 = 0; i < nums.size(); i++) {
            heap.push(nums[i]);
        }
        
        std::vector<int> result;
        for (int i = 0; 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)