用汇编来写一个Leetcode的题目

#include <stdio.h>

int TrappingResult(int height[],int len);           //总接水量
int CountTemp(int temp[],int len);                  //每行可接水量
int ArrMax(int height[],int len);                   //返回数组元素最大值
int main()
{
    int height[] = {0,1,0,2,1,0,1,3,2,1,2,1};
    int len = sizeof(height)/sizeof(int);           //数组长度
    int result = TrappingResult(height,len);        //结果
    printf("%d",result);
    return 0;
}

int TrappingResult(int height[],int len)
{
    int temp[len];
    int count = 0;
    int max = ArrMax(height,len);               //减少不必要的循环
    for(int i = 0;i < max;i ++)
        {
            for(int j = 0;j < len;j ++)
                {
                    temp[j] = height[j] > i?1:0;    //用循环化整为单，逐步计算
                }
            count+=CountTemp(temp,len);             //单个计算
        }
    return count;
}

int CountTemp(int temp[],int len)
{
    int head = 0,rear = -1;
    int count = 0;
    while(head < len-2)                 //当head指针大于数组长度-2时，不可能接到雨水
        {
            for(head = rear+1;head < len-1;head++)  //rear初始化为-1为了方便第一次循环
                {
                    if(temp[head] == 1)             //找到柱子
                        {
                            break;
                        }
                }
            for(rear = head;rear < len;rear++)      //寻找第二个柱子
                {
                    if(temp[rear+1] == 1)
                        {
                            count +=rear-head;      //计算接水量
                            break;
                        }
                }
        }
        return count;
}

int ArrMax(int height[],int len)
{
    int max=0;
    for(int i = 0;i < len;i ++)
        {
            if(height[i] > max)    max = height[i];
        }
    return max;
}