leetcode Largest Rectangle in Histogram

求解最大矩形，很容易想到O（n^2）的解法，只需要求解max(min(height(i...j)*(j-i+1))),0<=i<=j<n;

class Solution {public:    int largestRectangleArea(vector<int> &height) {         int ans=0,tans=0,len=height.size();        for(int i=0;i<len;i++){            int minh=height[i];            for(int j=0;j<=i;j++){                if(height[j]<minh)minh=height[j];                tans=(j-i+1)*minh;                if(tans>ans)ans=tans;            }        }        return ans;    }

但是对于大的数据超时，通过用栈记录记录，从而将空间复杂度由O(1)到O(n),以空间换时间，将时间复杂度降到O(n)

class Solution {public:    int largestRectangleArea(vector<int> &height) {                int len=height.size();        if(len==0)return 0;        stack<int >st;        stack<int >index;        int ans=0,tans=0;        for(int i=0;i<len;i++){            if(st.empty()||height[i]>st.top()){//递增的话，放入                st.push(height[i]);                index.push(i);            }else if(st.top()>height[i]){//出现降序                   int tindex=0;                   while(!st.empty()&&height[i]<st.top()){//计算山头部分的值                       tindex=index.top();                       index.pop();                       tans=st.top()*(i-tindex);                       st.pop();                       if(tans>ans)ans=tans;//记录最大                   }                   st.push(height[i]);//记录                   index.push(tindex);            }                        }        while(!st.empty()){//最后计算各个极小值情况               int tindex=index.top();               index.pop();               tans=st.top()*(len-tindex);               st.pop();               if(tans>ans)ans=tans;                   }        return ans;    }};

                                             
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