[LeetCode] 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 shad

For example, Given height =

[2,1,5,6,2,3]

, return

10

.

O（n）的方法：http://www.geeksforgeeks.org/largest-rectangle-under-histogram/

思路：借助栈的push和pop，依次把height里极高点的maxS求出来。

如果没有遇到极高点，就把height的序号push进栈；

如果遇到极高点，把极高点的maxS求出来，把极高点pop出栈；

依次类推！ 具体程序见下面代码：

class Solution {
public:
int largestRectangleArea(vector<int> &height) {
int maxS = 0;
stack<int> s;
int currS, current,low;
int len = height.size();
for(int i=0;i<len;)
{
if(s.empty() || height[i]>=height[s.top()])
s.push(i++);

else
{
current = s.top();
s.pop();
low = s.empty()?-1:s.top();
currS = height[current]*(i-low-1);
if(currS>maxS)
maxS = currS;
}//end while
}//end for
while(!s.empty())//此时s中剩下的是逐渐上升的矩形
{
current = s.top();
s.pop();
low = s.empty()?-1:s.top();
currS = height[current]*(len - low - 1);
if(currS>maxS)
maxS = currS;
}
return maxS;
}
};

下面的代码，和上面的基本思想一致，但是复杂度特别高TimeOut，感受一下：

class Solution {
public:
int largestRectangleArea(vector<int> &height) {
int len = height.size();
int maxS = 0;
for(int i=0;i<len;i++)
{
int newS = CentS(i,height);
if(maxS<newS)
maxS = newS;

}
return maxS;
}
private:
int CentS(int i,vector<int> &height)
{
int maxS = height[i];
int small=i-1,big = i+1;
while(small>=0 && height[small]>=height[i])
{
maxS += height[i];
small--;

}
while(big<height.size() && height[big]>=height[i])
{
maxS += height[i];
big++;
}

return maxS;
}
};

ed area, which has area =

10

unit.