Leetcode 215: Find Kth Largest Element in an array

题目：

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.

Example 1:

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

Example 2:

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

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

使用最小堆来解。创建一个大小为k的最小堆，把数组里的元素push进去。

#include 

int * g_queue;
int g_size;

void init( int k)
{
	g_queue = malloc(k * sizeof(int));
	g_size = 0;
}

void push(int d)
{
	int i = g_size++;
	while( i > 0)
	{
		int p = (i-1)/2;
		if(g_queue[p] <= d)
		{
			break;
		}

		g_queue[i] = g_queue[p];
		i = p;
	}

	g_queue[i] = d;
        return ret;
}


int pop(void)
{
	int ret = g_queue[0];
	int check = g_queue[--g_size];
	int i = 0;
	while( 2 * i + 1 < g_size)
	{
		int a = 2 * i + 1;
		int b = 2 * i + 2;
		
		if(b < g_size && g_queue[b] < g_queue[a])
		{
			a = b;
		}

	
		if(g_queue[a] >= check)
		{
			break;
		}

		g_queue[i] = g_queue[a];
		i = a;
	}

	g_queue[i] = check;
}

int findKthLargest(int* nums, int numsSize, int k){
	
	init(k);

	int i;
	for(i = 0; i < k; ++i)
	{
		push(nums[i]);
	}

	for(i = k; i < numsSize; ++i)
	{
		if(g_queue[0] < nums[i])
		{
			pop();
			push(nums[i]);			
		}
	}

	return g_queue[0];

}

int main(void)
{
	int nums[] = {3,2,1,5,6,4};
	int ret = findKthLargest(nums, 6, 2);

	return 0;
}

改进：

创建一个  k+1的堆，当堆的大小超过k的时候，pop。

int findKthLargest(int* nums, int numsSize, int k){
	
	init(k+1);

	int i;
	for(i = 0; i < numsSize; ++i)
	{
		push(nums[i]);
		if(g_size > k)
		{
			pop();
		}
	}

/*	for(i = k; i < numsSize; ++i)
	{
		if(g_queue[0] < nums[i])
		{
			pop();
			push(nums[i]);			
		}
	}
*/
	return g_queue[0];

}