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）
1 class Solution { 2 public: 3 int largestRectangleArea(vector<int> &height) { 4 int maxArea=0; 5 for(unsigned i=0; i<height.size(); i++) 6 { 7 int min = height[i]; 8 for(unsigned j=i; j<height.size(); j++) 9 { 10 if(height[j]<min) 11 min = height[j]; 12 int area = min*(j-i+1); 13 if(area>maxArea) 14 maxArea = area; 15 } 16 } 17 return maxArea; 18 } 19 };
方法二：用堆栈保存重要位置，复杂度O（n）
1 class Solution { 2 public: 3 int largestRectangleArea(vector<int> &height) { //用堆栈来实现 4 stack<unsigned> st; 5 unsigned maxArea = 0; 6 for(unsigned i=0; i<height.size(); i++) 7 { 8 if(st.empty()) 9 st.push(i); 10 else 11 { 12 while(!st.empty()) 13 { 14 if(height[i]>=height[st.top()]) 15 { 16 st.push(i); 17 break; 18 } 19 else 20 { 21 unsigned idx=st.top(); 22 st.pop(); 23 unsigned leftwidth = st.empty() ? idx : (idx-st.top()-1); 24 unsigned rightwidth = i - idx-1; 25 maxArea = max(maxArea, height[idx]*(leftwidth+rightwidth+1)); 26 } 27 } 28 if(st.empty()) 29 st.push(i); 30 } 31 } 32 unsigned rightidx = height.size(); 33 while(!st.empty()) 34 { 35 unsigned idx = st.top(); 36 st.pop(); 37 unsigned leftwidth = st.empty() ? idx : (idx-st.top()-1); 38 unsigned rightwidth = rightidx - idx-1; 39 maxArea = max(maxArea, height[idx]*(leftwidth+rightwidth+1)); 40 } 41 return maxArea; 42 } 43 };