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.

思路：

一个比较直观的思路是对于每个直方，分别向左右找到比它小的直方，然后求得从当前位置扩展可得到的最大值。然而这需要O（n^2）的复杂度，其中有很多重复的判断。

可以使用一个栈来实现上述思路。当栈为空或者栈顶元素小于当前值时，把当前值压入栈中，否则的话相当于找到了右边界，再从栈中弹出元素，因为栈内元素是递增的，所以弹出来的相当于是左边界。逐个计算面积，更新当前的max值，直到栈顶元素大于当前元素为止。弹栈时需要注意栈为空的情况。

代码：

 1     int max(int a, int b){
 2         if(a > b)
 3             return a;
 4         return b;
 5     }
 6     int largestRectangleArea(vector<int> &height) {
 7         // IMPORTANT: Please reset any member data you declared, as
 8         // the same Solution instance will be reused for each test case.
 9         stack<int> Stack;
10         int size = height.size();
11         if(size == 0)
12             return 0;
13         int i = 0;
14         int maxRec = 0;
15         while(i < size){
16             if(Stack.empty() || height[Stack.top()] <= height[i]){
17                 Stack.push(i++);
18             }
19             else{
20                 int t = Stack.top();
21                 Stack.pop();
22                 int tmpRec = height[t]*(Stack.empty()?i:i-Stack.top()-1);
23                 if(tmpRec > maxRec)
24                     maxRec = tmpRec;
25             }
26         }
27         while(!Stack.empty()){
28             int t = Stack.top();
29             Stack.pop();
30             int tmpRec = height[t]*(Stack.empty()?i:i-Stack.top()-1);
31             if(tmpRec > maxRec)
32                 maxRec = tmpRec;
33         }
34         return maxRec;
35     }

 

转载于:https://www.cnblogs.com/waruzhi/p/3439218.html