【LeetCode】215. Kth Largest Element in an Array

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215. Kth Largest Element in an Array

Description：
Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.
Difficulty：Medium

Example:

Input: [3,2,1,5,6,4] and k = 2
Output: 5

方法1：sort

• Time complexity : $O\left ( nlogn\right )$
• Space complexity : $O\left ( 1 \right )$

class Solution {
public:
	int findKthLargest(vector<int>& nums, int k) {
		sort(nums.begin(), nums.end(), [](const int& a, const int& b) {return a > b; });
		return nums[k - 1];
	}
};

方法2：min-heap

• Time complexity : $O\left ( nlogk\right )$
• Space complexity : $O\left ( k \right )$
  思路：
  维护k大小的最小堆，到最后的top便是结果

class Solution {
public:
	int findKthLargest(vector<int>& nums, int k) {
		priority_queue<int, vector<int>, greater<int> > pq;
		for (auto num : nums) {
			pq.push(num);
			if (pq.size() > k)
				pq.pop();
		}
		return pq.top();
	}
};

方法3：Quick select

• Time complexity : $O\left ( n\right )$
• Space complexity : $O\left ( 1 \right )$
  思路：
  利用快排中的partition，比快排要少一半的操作，因为每次左右两边的递归只运行一个即可。

class Solution {
public:
	int findKthLargest(vector<int>& nums, int k) {
        int left = 0, right = nums.size() - 1;
        while (true) {
            int p = partition(nums, left, right);
            if (p == k - 1) {
                return nums[p];
            }
            if (p > k - 1) {
                right = p - 1;
            } else {
                left = p + 1;
            }
        }
    }
    
    int partition(vector<int>& nums, int left, int right){
        int pivot = nums[left], l = left+1, r = right;
        while(true){
            while(l <= right && nums[l] > pivot) l++;
            while(left+1 <= r && nums[r] < pivot) r--;       
            if(l > r) break;
            swap(nums[l++], nums[r--]);
            }
        swap(nums[left], nums[r]);
        return r;
    }
};