leetcode 85. Maximal Rectangle

Posted 2020-10-10 =w=

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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.

题意：求出01矩阵中由1组成的最大的矩形面积。

思路：

哈哈，上一题的扩展版本。

对每个位置算出其往下几行连续的1的数量。然后就变成了上一题~

 

code:

class Solution {
public:
    int maximalRectangle(vector<vector<char>>& matrix) {
        if(matrix.size() == 0 || matrix[0].size() == 0) return 0;
        int n = matrix.size(), m = matrix[0].size();
        vector<vector<int>> state(n , vector<int>(m, 0));
        for(int i = 0; i < m; i++){
            state[n-1][i] = (matrix[n-1][i] == ‘1‘);
        }
        for(int i = n - 2; i >= 0; i --){
            for(int j = 0; j < m; j++){
                if(matrix[i][j] == ‘0‘) state[i][j] = 0;
                else{
                    state[i][j] = state[i+1][j] + 1;
                }
            }
        }
        
        int ans = 0;
        for(int i = 0; i < n; i++){
            stack<int>s;
            vector<int>leftPos(m);
            for(int j = 0; j < m; j++){
                while(!s.empty() && state[i][s.top()] >= state[i][j]) s.pop();
                if(s.empty()) leftPos[j] = 0;
                else{
                    leftPos[j] = s.top() + 1;
                }
                s.push(j);
            }
            while(!s.empty()) s.pop();
            for(int j = m - 1; j >= 0; j--){
                while(!s.empty() && state[i][s.top()] >= state[i][j] ) s.pop();
                int rightPos = m - 1;
                if(!s.empty()) rightPos = s.top() - 1;
                int tmp = (rightPos - leftPos[j] + 1) * state[i][j];
                ans = max(ans, tmp);
                s.push(j);
            }
        }
        return ans;
        
    }
};