47. Largest Rectangle in Histogram && Maximal Rectangle

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<int> &height) {
        int n = height.size(), max_area = 0;
        int Id, area;
        int i = 0;
        stack<int> 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<int> &h) {
    stack<int> 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<vector<char> > &matrix) {
        if(matrix.size() == 0 || matrix[0].size() == 0) return 0;
        int row = matrix.size(), col = matrix[0].size();
        vector<int> 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;
    }
};