用c语言手写堆排序以及来道简单题吧 1046. 最后一块石头的重量

堆排序手写一遍吧

#include<stdio.h>
#include<stdlib.h>
void swap(int *a,int *b)
{
	int temp=*a;
	*a=*b;
	*b=temp;
 } 
void HeapAdjust(int *heap,int i,int size)
{
	int lchild=2*i+1;//左孩子序号 
	int rchild=2*i+2;//右孩子序号
	int temp =i;
	if(i<=size/2-1)
	{
		if(lchild<=size-1&&heap[lchild]>heap[temp])
		{
			temp=lchild;
		}
		if(rchild<=size-1&&heap[rchild]>heap[temp])
		{
			temp=rchild;
		}
		if(temp!=i)
		{
			swap(&heap[i],&heap[temp]);
			HeapAdjust(heap,temp,size);//避免交换后的子树不满足堆 
		}
	}	
} 

void BuildHeap(int *heap,int size)
{
	int i;
	for(i=size/2-1;i>=0;i--)
	{
		HeapAdjust(heap,i,size);
	}	
}

void HeapSort(int *heap,int size)
{
	int i;
	BuildHeap(heap,size);
	for(i=size-1;i>=0;i--)
	{
		swap(&heap[0],&heap[i]);//每次将最大的丢在数组末尾，实现排序 
		HeapAdjust(heap,0,i-1);//将剩余结点重新调整成大顶堆 
	}
}

Leetcode例题 1046. 最后一块石头的重量

解法

void swap(int *a,int *b)
{
	int temp=*a;
	*a=*b;
	*b=temp;
} 
void HeapAdjust(int *heap, int i,int size)
{

    int lchild=2*i+1;//左孩子序号 
	int rchild=2*i+2;//右孩子序号
	int temp =i;
	if(i<=size/2-1)
	{
		if(lchild<=size-1&&heap[lchild]>heap[temp])
		{
			temp=lchild;
		}
		if(rchild<=size-1&&heap[rchild]>heap[temp])
		{
			temp=rchild;
		}
		if(temp!=i)
		{
			swap(&heap[i],&heap[temp]);
			HeapAdjust(heap,temp,size);//避免交换后的子树不满足堆 
		}
	}	
}

void buildHeap(int *heap, int size) {
    for (int i=size-1; i>=0; --i) {
        HeapAdjust(heap, i, size);
    }
}

int fetch(int *heap, int *size) {
    int x = heap[0];
    heap[0] = heap[--(*size)];
    HeapAdjust(heap, 0, *size);
    return x;
}

void push(int *heap, int *size, int x) {
    heap[(*size)++] = x;
    HeapAdjust(heap, *size-1, *size);
}

int lastStoneWeight(int* stones, int stonesSize){
    buildHeap(stones, stonesSize);
    while (stonesSize > 1) {
        int x = fetch(stones, &stonesSize);
        int y = fetch(stones, &stonesSize);
        if (x == y) {
            // do nothing
        } else {
            int t = abs(x-y);
            push(stones, &stonesSize, t);
        }
    }
    return stonesSize ? stones[0] : 0;
}

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