problem：
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.
thinking：
（1）最先想到暴力破解，两次遍历，记录最大面积。提交发现超时了。
（2）http://blog.csdn.net/doc_sgl/article/details/11805519 给出了一个用stack实现的O（n）算法，很高效。
大致思想是：stack里面只存放height单调递增的索引，最大矩形最可能出现在这几列中
code：
暴力破解： 提交 TLE
class Solution {
public:
int largestRectangleArea(vector<int> &height) {
int n=height.size();
if(n==0)
return 0;
if(n<2)
return height[0];
int area=height[0];
int index=0;
for(int i=1;i<n;i++)
{
index=height[i];
for(int j=i;j>=0;j--)
{
index=min(index,height[j]);
area=max(area,index*(i-j+1));
}
}
return area;
}
};借助stack O（N）算法
class Solution {
public:
int Max(int a, int b){return a > b ? a : b;}
int largestRectangleArea(vector<int> &height) {
height.push_back(0);
stack<int> stk;
int i = 0;
int maxArea = 0;
while(i < height.size()){
if(stk.empty() || height[stk.top()] <= height[i]){
stk.push(i++);
}else {
int t = stk.top();
stk.pop();
maxArea = Max(maxArea, height[t] * (stk.empty() ? i : i - stk.top() - 1));
}
}
return maxArea;
}
};leetcode || 84、Largest Rectangle in Histogram
原文地址：http://blog.csdn.net/hustyangju/article/details/45042517
