最大矩形-LintCode

给你一个二维矩阵，权值为False和True，找到一个最大的矩形，使得里面的值全部为True，输出它的面积

样例：
给你一个矩阵如下

[  [1, 1, 0, 0, 1],  [0, 1, 0, 0, 1],  [0, 0, 1, 1, 1],  [0, 0, 1, 1, 1],  [0, 0, 0, 0, 1]]

输出6

思路：
参考题目“直方图最大矩形覆盖”，将矩阵的每一行当做是直方图来求解。

#ifndef C510_H#define C510_H#include<iostream>#include<vector>#include<stack>using namespace std;class Solution {public:    /*    * @param matrix: a boolean 2D matrix    * @return: an integer    */    int maximalRectangle(vector<vector<bool>> &matrix) {        // write your code here        if (matrix.empty())            return 0;        int row = matrix.size();        int col = matrix[0].size();        vector<vector<int>> v(row, vector<int>(col, 0));        for (int j = 0; j < col; ++j)        {            for (int i = 0; i < row; ++i)            {                if (matrix[i][j])                {                    v[i][j] = 1;                    if (i>0)                        v[i][j] += v[i - 1][j];                }                       }        }        int area = 0;        for (auto &c : v)        {            area = maxVal(largestRectangleArea(c), area);        }        return area;    }    int largestRectangleArea(vector<int> &height) {        int res = 0;        stack<int> s;        height.push_back(0);        for (int i = 0; i < height.size(); ++i) {            if (s.empty() || height[s.top()] < height[i]) s.push(i);            else {                int cur = s.top();                s.pop();                res = maxVal(res, height[cur] * (s.empty() ? i : (i - s.top() - 1)));                --i;            }        }        return res;    }    int maxVal(int a, int b)    {        return a > b ? a : b;    }};#endif