84. Largest Rectangle in Histogram *HARD* 柱状图求最大面积 85. Maximal Rectangle *HARD*

1.

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 heights = [2,1,5,6,2,3],
return 10.

class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        int n = heights.size();
        if(0 == n)
            return 0;
        int max = 0, area, i, k;
        stack<int> s;
        heights.push_back(0);
        for(i = 0; i <= n; i++)
        {
            if(s.empty() || heights[i] >= heights[s.top()])
            {
                s.push(i);
                continue;
            }
            k = s.top();
            s.pop();
            area = heights[k] * (0 == s.size() ? i : i - s.top() - 1);
            if(area > max)
                max = area;
            i--;
        }
        return max;
    }
};

// As we know, the area = width * height
// For every bar, the ‘height‘ is determined by the loweset bar.
//
// 1) We traverse all bars from left to right, maintain a stack of bars. Every bar is pushed to stack once.
// 2) A bar is popped from stack when a bar of smaller height is seen.
// 3) When a bar is popped, we calculate the area with the popped bar as smallest bar.
// 4) How do we get left and right indexes of the popped bar –
// the current index tells us the ‘right index’ and index of previous item in stack is the ‘left index’.
//
//
// In other word, the stack only stores the incresing bars, let‘s see some example
//
// Example 1
// ---------
// height = [1,2,3,4]
//
// stack[] = [ 0, 1, 2, 3 ], i=4
//
// 1) pop 3, area = height[3] * 1 = 4
// 2) pop 2, area = height[2] * 2 = 4
// 3) pop 1, area = height[1] * 3 = 6
// 4) pop 0, area = height[0] * 4 = 4
//
//
// Example 2
// ---------
// height = [2,1,2]
//
// stack[] = [ 0 ], i=1
// 1) pop 0, area = height[0] * 1 = 2
//
// stack[] = [ 1,2 ], i=3, meet the end
// 1) pop 2, area = height[2] * 1 = 2
// 2) pop 1, area = height[1] * 3 = 3
//
//
// Example 3
// ---------
// height = [4,2,0,3,2,5]
//
// stack[] = [ 0 ], i=1, height[1] goes down
// 1) pop 0, area = height[0] * 1 = 4
//
// stack[] = [ 1 ], i=2, height[2] goes down
// 1) pop 1, area = height[1] * 2 = 4 // <- how do we know the left?
// start from the 0 ??
//
// stack[] = [ 2, 3 ], i=4, height[4] goes down
// 1) pop 3, area = height[3] * 1 = 3
// 2) pop 2, area = height[2] * ? = 0 // <- how do we know the left?
// start from the 0 ??
//
// stack[] = [ 2,4,5 ], i=6, meet the end
// 1) pop 5, area = height[5] * 1 = 5
// 2) pop 4, area = height[4] * 3 = 6 // <- how do we know the left?
// need check the previous item.
// 3) pop 2, area = height[2] * ? = 4 // <- how do we know the left?
// start from the 0 ??
//
// so, we can see, when the stack pop the top, the area formular is
//
// height[stack_pop] * i - stack[current_top] - 1, if stack is not empty
// height[stack_pop] * i, if stack is empty

2.

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

class Solution {
public:
    int maxRecArea(vector<int> heights)
    {
        stack<int> s;
        int n = heights.size(), max = 0, area, i, k;
        heights.push_back(0);
        for(i = 0; i <= n; i++)
        {
            if(s.empty() || heights[i] >= heights[s.top()])
            {
                s.push(i);
                continue;
            }
            k = s.top();
            s.pop();
            area = heights[k] * (s.empty() ? i : i - s.top() - 1);
            if(area > max)
                max = area;
            i--;
        }
        return max;
    }
    
    int maximalRectangle(vector<vector<char>>& matrix) {
        int m = matrix.size();
        if(0 == m)
            return 0;
        int n = matrix[0].size(), area, max = 0, i, j;
        vector<vector<int>> heights(m, vector<int>(n, 0));
        for(i = 0; i < m; i++)
        {
            for(j = 0; j < n; j++)
            {
                if(‘1‘ == matrix[i][j])
                    heights[i][j] = (0 == i ? 1 : heights[i-1][j]+1);
            }
            area = maxRecArea(heights[i]);
            if(area > max)
                max = area;
        }
        return max;
    }
};

// The problem can be convert to the problem - "Largest Rectangle in Histogram"
// 1) we can take each row to calculate each row‘s histogram.
// 2) using the algorithm of "Largest Rectangle in Histogram" to find the largest area histogram.
// 3) tracking the maximal area.
//
// For the 1), it‘s easy.
// heights[i][j] = 1, if (i==0)
// heights[i][j] = heights[i-1][j] + 1;, if (i>0)
//
// For the 2), please referr to "Largest Rectangle in Histogram"

84. Largest Rectangle in Histogram *HARD* 柱状图求最大面积 85. Maximal Rectangle *HARD*

原文：http://www.cnblogs.com/argenbarbie/p/5338834.html