LeetcodeLargest -84- Rectangle in Histogram 94.01%

今天第一天开始刷Leetcode，一不小心抽到了一道hard，搞了一下午也没搞出来，半夜突然有了灵感，想用递归试一试，结果调了调居然AC了，效率还高于94.01%·······
先把代码贴在下面了，图解什么的······该休息了，改天吧

class Solution {public:    int largestRectangleArea(vector<int>& heights) {        if ((heights.size() == 0))            return 0;       //判断输入不为空        vector<int>::iterator it = heights.begin(), it_end=heights.end();        int max = 0, remain=0;        while (it != heights.end())            //跳出递归的两种条件：it迭代至end、当前项小于最底层递归的高度标记            //第二种情况需要将上次递归中返回的count作为rest带入下一次递归            remain = recursion_84(it, it_end, max, remain   );        return max;    }    int recursion_84(vector<int>::iterator &it, vector<int>::iterator &it_end, int &max, int past)    {        int count = past, height = *it, remain=0, temp=0;//进入函数时it为第一项，本层递归的识别标志为height        while (height!=0)//主循环        {            while ((it != it_end) && (height == *it))//将与该层标志相等的项目累加至count                count++, it++, remain=0;            if ((it == it_end)||(height > *it))//迭代至尾部，或下一层低于本层标记时->判断是否为max，并结束本层递归            {                if (max < count*height)                    max = count*height;                return count;            }            else//下一项高于本层标记，进入下一层递归            {                //下层递归结束后返回下层递归的count累加至本层count之上                //将past存入remain，然后继续主循环                temp = recursion_84(it, it_end, max, remain);                 count +=temp-remain;                remain = temp;            }        }        it++;        return 0;    }};