LeetCode 215. Kth Largest Element in an Array

最新推荐文章于 2023-12-15 19:39:48 发布

原创最新推荐文章于 2023-12-15 19:39:48 发布 · 227 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 题解专栏收录该内容

288 篇文章

订阅专栏

本文介绍了一种方法来找到未排序数组中第K大的元素，通过使用优先队列实现，同时分析了时间复杂度和空间复杂度。

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.

Note:
You may assume k is always valid, 1 ≤ k ≤ array's length.

#include <vector>
#include <iostream>
#include <queue>
using namespace std;

// This method is easy but need O(n) memory space and O(n) time complexity.
int findKthUsingQueue(vector<int>& nums, int k) {
    // kth largest equals to (nums.size() - k + 1)th smallest.
    priority_queue<int, vector<int>, greater<int> > numsQueue;
    for(int i = 0; i < nums.size(); ++i) {
        numsQueue.push(nums[i]);
    }

    for(int i = 1;  i < nums.size() - k + 1; i++) {
        numsQueue.pop();
    }
    return numsQueue.top();
}
int partition(vector<int>& nums, int left, int right);

// it currently find kth-1 smallest number.
int findKth(vector<int>& nums, int first, int last, int k) {
    int pivot = partition(nums, first, last);
    if(pivot == k) return nums[pivot];
    else if(pivot < k) return findKth(nums, pivot + 1, last, k);
    else return findKth(nums, first, pivot - 1, k);
}

// O(n) time complexity and O(1) space.
int partition(vector<int>& nums, int left, int right) {
    int pivotElement = nums[left];
    int pivotIndex = left;
    for(int i = left + 1; i <= right; ++i) {
        if(nums[i] < pivotElement) {
            pivotIndex++;
            swap(nums[i], nums[pivotIndex]);
        }
    }
    swap(nums[pivotIndex], nums[left]);
    // the index p right now point to the pivot.
    return pivotIndex;
}
 int main(void) {
    vector<int> nums{4, 3, 2, 5, 8, 1};
    int res = findKthUsingQueue(nums, 2);
    int res1 = findKth(nums, 0, 5, 5);
    cout << res1 << endl;
}