85. Maximal Rectangle

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

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Return 6.

解法1:看了半天的discuss，不得不说动态规划的思想真的很难搞。看懂之后就觉得也没想象那么难，但是为什么自己想不到呢，还是因为做题不多的原因吧，作者链接:https://discuss.leetcode.com/topic/6650/share-my-dp-solution

计算面积的公式：(right[i,j]-left[i,j])*height[i,j] 左右边界是左闭右开的。

left(i,j) = max(left(i-1,j), cur_left), cur_left can be determined from the current row   //left 主要是记录当前高度下最左的边

right(i,j) = min(right(i-1,j), cur_right), cur_right can be determined from the current row  //right主要是记录当前高度最右边+1

height(i,j) = height(i-1,j) + 1, if matrix[i][j]=='1';  //height记录从上往下连续的1的个数

height(i,j) = 0, if matrix[i][j]=='0' 。  

class Solution {
public:
    int maximalRectangle(vector<vector<char>>& matrix) {
        if(matrix.empty())return 0;
        int n=matrix.size();
        int m=matrix[0].size();
        int left[m],right[m],height[m],ans=0;
       // fill_n(left,m,0),fill_n(right,m,m),fill_n(right,m,0);
        fill(left,left+m,0),fill(right,right+m,m),fill(height,height+m,0);
        for(int i=0;i<n;i++){
            int cur_left=0,cur_right=m;
            for(int j=0;j<m;j++){
                if(matrix[i][j]=='0')height[j]=0;
                else height[j]++;
            }
            for(int j=0;j<m;j++){
                if(matrix[i][j]=='0')cur_left=j+1,left[j]=0;
                else left[j]=max(left[j],cur_left);
            }
            for(int j=m-1;j>=0;j--){
                if(matrix[i][j]=='0')cur_right=j,right[j]=m;
                else right[j]=min(right[j],cur_right);
            }
            for(int j=0;j<m;j++){
              //  printf("left:%d right:%d height:%d\n",left[j],right[j],height[j]);
                ans=max(ans,(right[j]-left[j])*height[j]);
            }
        }
        return ans;
    }
};

 解法2:参照Largest Rectangle in Histogram的思路，前者是计算给定高度排列的最大矩形面积，此处只要每行遍历过去，从该行往上连续的1就是前者题目的高度。

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