[leetcode] Kth Largest Element in an Array

最新推荐文章于 2024-11-19 02:25:22 发布

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本文介绍了一种高效的方法来查找未排序数组中的第k大元素，通过使用多集数据结构实现，确保了算法的时间复杂度为O(n log k)，适用于实际应用场景。

From : https://leetcode.com/problems/kth-largest-element-in-an-array/

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.

For example,
Given [3,2,1,5,6,4] and k = 2, return 5.

class Solution {
public:
    int findKthLargest(vector<int>& nums, int k) {
        if(nums.size()<k || k<=0) return 0;
		multiset<int, less<int>> box;
		multiset<int, less<int>>::iterator it;
        for(int i=0; i<nums.size(); i++) {
            if(i>=k) {
				it = box.begin();
                if(*it<nums[i]) {
                    box.erase(it);
                    box.insert(nums[i]);
                }
            } else box.insert(nums[i]);
        }
        return *box.begin();
    }
};

public class Solution {
    public int findKthLargest(int[] nums, int k) {
        int len = nums.length, left=0, right = len-1, index=partation(nums, left, right);
        while(len != index+k) {
            if(index+k > len) {
                right = index-1;
            } else {
                left = index+1;
            }
            index=partation(nums, left, right);
        }
        return nums[index];
    }
    private int partation(int[] nums, int left, int right) {
    	int pos = left, v = nums[pos];
    	while(left <= right) {
    		while(left<=right && nums[left]<=v) ++left;
    		while(left<=right && nums[right]>v) --right;
    		if(left < right) {
    			swap(nums, left, right);
    		}
    	}
    	swap(nums, pos, right);
    	return right;
    }
    
    void swap(int[] nums, int i, int j) {
    	int t = nums[i];
    	nums[i] = nums[j];
    	nums[j] = t;
    }
}

public class Solution {
    public int findKthLargest(int[] nums, int k) {
        PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
        int len = nums.length;
        for(int i=0; i < len && i < k; ++i) {
        	pq.add(nums[i]);
        }
        for(int i=k; i<len; ++i) {
        	if(pq.peek() < nums[i]) {
        		pq.poll();
        		pq.add(nums[i]);
        	}
        }        
        return pq.peek();
    }
}