leetcode[84]Largest Rectangle in Histogram

最新推荐文章于 2020-09-08 05:21:43 发布

转载最新推荐文章于 2020-09-08 05:21:43 发布 · 42 阅读

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原文链接：http://www.cnblogs.com/Vae1990Silence/p/4281413.html

本文介绍了一种求解最大直方图面积的高效算法。该算法通过使用栈来跟踪直方图中柱子的高度，并计算出能构成的最大矩形面积。通过具体的代码实现和示例说明了算法的工作原理。

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 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 - t));  
            }  
        }  
        return maxArea;  
    }

int largestRectangleArea(vector<int> &height) 
{
    if(height.empty())return 0;
    height.push_back(0);
    int maxRes=0;
    stack<int> sta;
    for (int i=0;i<height.size();i++)
    {
        if (sta.empty()||(!sta.empty()&&height[i]>=height[sta.top()]))
        {
            sta.push(i);
        }
        else
        {
            int top;
            int newRes;
            while(!sta.empty()&&height[i]<height[sta.top()])
            {
                top=sta.top();
                sta.pop();
                newRes=(sta.empty()?i:(i-sta.top()-1))*height[top];
                maxRes=maxRes>newRes?maxRes:newRes;
                
            }
            sta.push(i);
        }
    }
    return maxRes;
}
*/
int largestRectangleArea(vector<int> &height) 
{
    if(height.empty())return 0;
    height.push_back(0);
    int maxRes=0;
    stack<int> sta;
    for (int i=0;i<height.size();i++)
    {
        int top;
        int newRes;
        while(!sta.empty()&&height[i]<height[sta.top()])
        {
            top=sta.top();
            sta.pop();
            newRes=(sta.empty()?i:(i-sta.top()-1))*height[top];
            maxRes=maxRes>newRes?maxRes:newRes;                
        }
        sta.push(i);
    }
    return maxRes;
}
};

转载于:https://www.cnblogs.com/Vae1990Silence/p/4281413.html