[LeetCode OJ] Largest Rectangle in Histogram

Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].

The largest rectangle is shown in the shaded area, which has area = 10 unit.

For example,
Given height = [2,1,5,6,2,3],
return 10.

方法一：两层循环遍历，复杂度O（n²）

 class Solution {
 public:
     int largestRectangleArea(vector<int> &height) {
         int maxArea=;
         for(unsigned i=; i<height.size(); i++)
         {
             int min = height[i];
             for(unsigned j=i; j<height.size(); j++)
             {
                 if(height[j]<min)
                     min = height[j];
                 int area = min*(j-i+);
                 if(area>maxArea)
                     maxArea = area;
             }
         }
         return maxArea;
     }
 };

方法二：用堆栈保存重要位置，复杂度O（n）

 class Solution {
 public:
     int largestRectangleArea(vector<int> &height) {  //用堆栈来实现
         stack<unsigned> st;
         unsigned maxArea = ;
         for(unsigned i=; i<height.size(); i++)
         {
             if(st.empty())
                 st.push(i);
             else
             {
                 while(!st.empty())
                 {
                     if(height[i]>=height[st.top()])
                     {
                         st.push(i);
                         break;
                     }
                     else
                     {
                         unsigned idx=st.top();
                         st.pop();
                         unsigned leftwidth = st.empty() ? idx : (idx-st.top()-);
                         unsigned rightwidth = i - idx-;
                         maxArea = max(maxArea, height[idx]*(leftwidth+rightwidth+));
                     }
                 }
                 if(st.empty())
                     st.push(i);
             }
         }
         unsigned rightidx = height.size();
         while(!st.empty())
         {
             unsigned idx = st.top();
             st.pop();
             unsigned leftwidth = st.empty() ? idx : (idx-st.top()-);
             unsigned rightwidth = rightidx - idx-;
             maxArea = max(maxArea, height[idx]*(leftwidth+rightwidth+));
         }
         return maxArea;
     }
 };