Leetcode:85. Maximal Rectangle

Difficulty: Hard
Ac 44ms

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

这道题一开始并没有什么好的思路，只能先截取每行为底，对每列求高，简化成一列数组球最大面积的问题，这样的复杂度大概是o(n*n)
如

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

可以

0 2 1 3
0 1 0 2
0 0 0 1

数组求最大面积的问题

class Solution {
public:
    int maximalRectangle(vector<vector<char> > &matrix) {
    if(matrix.empty()){
        return 0;
    }
    int maxRec = 0;
    vector<int> height(matrix[0].size(), 0);
    for(int i = 0; i < matrix.size(); i++){
        for(int j = 0; j < matrix[0].size(); j++){
            if(matrix[i][j] == '0'){
                height[j] = 0;
            }
            else{
                height[j]++;
            }
        }
        maxRec = max(maxRec, largestRectangleArea(height));
    }
    return maxRec;
}

int largestRectangleArea(vector<int> &height) {
    stack<int> s;
    height.push_back(0);
    int maxSize = 0;
    for(int i = 0; i < height.size(); i++){
        if(s.empty() || height[i] >= height[s.top()]){
            s.push(i);
        }
        else{
            int temp = height[s.top()];
            s.pop();
            maxSize = max(maxSize, temp * (s.empty() ? i : i - 1 - s.top()));
            i--;
        }
    }
    return maxSize;
}
};

看到有人用动态规划写出来的，速度很快大概思路是这样的。

0 0 0 1 0 0 0 
0 0 1 1 1 0 0 
0 1 1 1 1 1 0
The vector "left" and "right" from row 0 to row 2 are as follows

row 0:

l: 0 0 0 3 0 0 0
r: 7 7 7 4 7 7 7
row 1:

l: 0 0 2 3 2 0 0
r: 7 7 5 4 5 7 7 
row 2:

l: 0 1 2 3 2 1 0
r: 7 6 5 4 5 6 7

class Solution {
    public:
int maximalRectangle(vector<vector<char> > &matrix) {
    if(matrix.empty()) return 0;
    const int m = matrix.size();
    const int n = matrix[0].size();
    int left[n], right[n], height[n];
    fill_n(left,n,0); fill_n(right,n,n); fill_n(height,n,0);
    int maxA = 0;
    for(int i=0; iint cur_left=0, cur_right=n; 
        for(int j=0; j// compute height (can do this from either side)
            if(matrix[i][j]=='1') height[j]++; 
            else height[j]=0;
        }
        for(int j=0; j// compute left (from left to right)
            if(matrix[i][j]=='1') left[j]=max(left[j],cur_left);
            else {left[j]=0; cur_left=j+1;}
        }
        // compute right (from right to left)
        for(int j=n-1; j>=0; j--) {
            if(matrix[i][j]=='1') right[j]=min(right[j],cur_right);
            else {right[j]=n; cur_right=j;}    
        }
        // compute the area of rectangle (can do this from either side)
        for(int j=0; jreturn maxA;
}
};