【leetcode】Kth Largest Element in an Array （middle）☆

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.

 

思路：

堆。讲解：二叉堆

class Solution {

public:

    //新插入i结点 其父节点为(i - 1) / 2

    void MinHeapFixup(int a[], int i) 

    {

        int j = (i - 1) / 2; //父节点

        int temp = a[i];

        while(j >= 0 && i != 0)

        {

            if(a[j] <= temp) break;

            a[i] = a[j];

            i = j;

            j = (i - 1) / 2;

        }

        a[i] = temp;

    }



    //在最小堆中插入新数据nNum

    void MinHeapAddNumber(int a[], int n, int nNum)

    {

        a[n] = nNum;

        MinHeapFixup(a, n);

    }



    //堆删除后的调整  从i节点开始调整,n为节点总数 从0开始计算 i节点的子节点为 2*i+1, 2*i+2

    void MinHeapFixdown(int a[], int i, int n)

    {

        int j = 2 * i + 1;

        int temp = a[i];

        while(j < n)

        {

            if(j + 1 < n && a[j + 1] < a[j]) //在左右孩子中找最小的

                j++;

            if(a[j] >= temp) 

                break;

            a[i] = a[j];  //把较小的子结点往上移动,替换它的父结点

            i = j;

            j = 2 * i  + 1;

        }

        a[i] = temp;

    }



    //在最小堆中删除数

    void MinHeapDeleteNumber(int a[], int n)

    {

        swap(a[0], a[n - 1]);

        MinHeapFixdown(a, 0, n - 1);

    }



    int findKthLargest(vector<int>& nums, int k) {

        int * a = new int[k]; //大小为k的最小堆

        for(int i = 0; i < nums.size(); ++i)

        {

            if(i < k)

            {

                MinHeapAddNumber(a, i, nums[i]); //插入数据

            }

            else if(nums[i] > a[0]) //比已有的k个最大的数字大

            {

                MinHeapDeleteNumber(a, k);

                MinHeapAddNumber(a, k - 1, nums[i]);

            }

        }

        return a[0];

    }

};

 

用STL的堆：

 int findKthLargest(vector<int>& nums, int k) {

    priority_queue<int> p;

    const int s(nums.size());



    for (int i = 0; i < s; ++i) p.push(nums[i]);

    while (--k) p.pop();



    return p.top();

}

堆的相关讲解：

http://www.cnblogs.com/flyoung2008/articles/2136485.html