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.
class Solution {
public:
int largestRectangleArea(vector<int> &height) {
int ans = 0,area;
stack<int> s;
height.push_back(0);
for(int i = 0;i < height.size();i ++){
if(s.empty() || !s.empty() && height[i] > height[s.top()])
s.push(i);
else{
while(!s.empty() && height[s.top()] > height[i]){
int idx = s.top();
s.pop();
int width = s.empty() ? i : i - s.top() - 1;
area = height[idx]*width;
ans = max(ans,area);
}
s.push(i);
}
}
return ans;
}
};
本文介绍了一种求解直方图中最大矩形面积的高效算法。通过使用栈来跟踪直方图条的高度,该算法能够在线性时间内找到最大矩形区域。举例说明了如何实现这一算法并给出了具体的代码实现。
904

被折叠的 条评论
为什么被折叠?



