LeetCode-maximal-rectangle(最大矩阵面积)

题目描述

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 =10unit.

For example,
Given height =[2,1,5,6,2,3],
return10.

算法思路：

请参考：https://blog.csdn.net/u012534831/article/details/74356851

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        int n=static_cast<int>(height.size());
        if(n<=0) return 0;
        int ans=0,num;
        stack<int>st;
        st.push(-1);
        for(int i=0;i<n;++i)
        {
            while(st.top()!=-1&&height[i]<height[st.top()])
            {
                num=st.top();
                st.pop();
                ans=max(ans,(i-st.top()-1)*height[num]);
            }
            st.push(i);
        }
        while(st.top()!=-1)
            {
                num=st.top();
                st.pop();
                ans=max(ans,(n-st.top()-1)*height[num]);
            }
        return ans;
    }
};

相同算法拓展：

题目描述

Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing all ones and return its area.

分别遍历每一行，将每次遍历的行及以上部分看做直方图求最大面积，最后遍历所有结果，求最大值。

class Solution {
public:
    int maximalRectangle(vector<vector<char> > &matrix) {
        int m=matrix.size();
        if(m<=0) return 0;
        int n=matrix[0].size();
        vector<int>h(n);
        fill(h.begin(),h.end(),0);
        int num;
        int ans=0;
        stack<int>st;
        st.push(-1);
        for(int i=0;i<m;++i)
        {
            for(int j=0;j<n;++j)
            {
                if(matrix[i][j]=='1')
                    ++h[j];
                else
                    h[j]=0;
            }
            for(int j=0;j<n;++j)
            {
                while(st.top()!=-1&&h[j]<h[st.top()])
                {
                    num=st.top();
                    st.pop();
                    ans=max(ans,(j-1-st.top())*h[num]);
                }
                st.push(j);
            }
            while(st.top()!=-1)
            {
                num=st.top();
                st.pop();
                ans=max(ans,(n-1-st.top())*h[num]);
            }    
        }
        return ans;
    }
};