求数组第K大

最新推荐文章于 2024-08-26 09:22:10 发布

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本文介绍了一种高效算法，用于从无序数组中找到第K大的元素。通过快速选择算法实现，该算法采用分区思想，随机选取支点，并进行一次遍历完成元素排序。这种方法不仅简单易懂，而且平均时间复杂度能达到O(n)。

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.

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

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class Solution
{
public:
    int findKthLargest(vector<int>& nums, int k)
    {
        int ans=  selectk(nums,0,nums.size()-1,k);
        return nums[ans];
    }
    int selectk(vector<int>& a, int low, int high, int k)
    {
        if(k <= 0)
            return -1;
        if(k > high - low + 1)
            return -1;
        int pivot = low + rand()%(high - low + 1);    //随即选择一个支点
        swap(a[low], a[pivot]);
        int m = low;
        int count = 1;

        //一趟遍历，把较大的数放到数组的左边

        for(int i = low + 1; i <= high; ++i)
        {
            if(a[i] > a[low])
            {
                swap(a[++m], a[i]);
                count++;              //比支点大的数的个数为count-1
            }
        }
        swap(a[m], a[low]);           //将支点放在左、右两部分的分界处
        if(count > k)
        {
            return selectk(a, low, m - 1, k);
        }
        else if( count < k)
        {
            return selectk(a, m + 1, high, k - count);
        }
        else
        {
            return m;
        }
    }
};