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： 注意一点: 只要计算出以每个柱形为最小值的矩形面积即可。使用一个栈，栈内始终保存一个递增序列的 index，若是 新的柱形长度小于栈顶元素，则退出栈顶直到栈内元素的长度不大于新的柱形的长度为止，并且，对于每一个退栈元素，计算以其长度为最小值的面积。（宽的左边为其自身位置，右边为新到元素的位置）时间：O(n)

class Solution {public:    int largestRectangleArea(vector
     
       &height) {        int n = height.size(), max_area = 0;        int Id, area;        int i = 0;        stack
      
        st;  // save the index        while(i < n) {            if(st.empty() || height[i] >= height[st.top()]) st.push(i++);            else {                Id = st.top();                st.pop();                area = height[Id] * (st.empty() ? i : i - st.top() - 1);                if(area > max_area) max_area = area;            }        }        while(!st.empty()) {            Id = st.top();            st.pop();            area = height[Id] * (st.empty() ? i : i - st.top() - 1);            if(area > max_area) max_area = area;        }        return max_area;    }};
      
     

 

Maximal Rectangle

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

思路： 一行一行的记录下当前高度, 用上题的思路计算一下，即可。时间： O(n²)

int getMaxArea(vector
     
       &h) {    stack
      
        st;    int maxArea, i = 0;    while(i < h.size()) {        if(st.empty() || h[st.top()] <= h[i]) { st.push(i++); continue; }        while(!st.empty() && h[st.top()] > h[i]) {            int id = st.top();            st.pop();            int area = h[id] * (st.empty() ? i : i - st.top() - 1);            if(area > maxArea) maxArea = area;        }    }    while(!st.empty()) {        int id = st.top();        st.pop();        int area = h[id] * (st.empty() ? i : i - st.top() - 1);        if(area > maxArea) maxArea = area;    }    return maxArea;}class Solution {public:    int maximalRectangle(vector
       
        
          > &matrix) {        if(matrix.size() == 0 || matrix[0].size() == 0) return 0;        int row = matrix.size(), col = matrix[0].size();        vector
         
           h(row, 0);        int maxArea = 0;        for(int c = 0; c < col; ++c) {            for(int r = 0; r < row; ++r) {                if(matrix[r][c] == '1')  ++h[r];                else h[r] = 0;            }            maxArea = max(maxArea, getMaxArea);        }        return maxArea;    }};