221.Maximal Square

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Maximal Square

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

Example:

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Input: 

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

Output: 4

解法1:单调栈

解法详见程序员代码面试指南：求最大子矩阵的大小

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class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) {
        if(matrix.empty() || matrix[0].empty()){
            return 0;
        }
        int maxArea = 0;
        int row = matrix.size(); // 1
        int col = matrix[0].size(); // 1
        vector<int> height(col, 0);
        for(int i=0;i<row;++i){
            
            for(int j=0;j<col;++j){
                height[j] = (matrix[i][j] == '0') ? 0 :  height[j] + 1;
            }
            auto res = getLessThan(height);
            for(int j=0;j<col;++j){
                int w = res[j][1] - res[j][0] - 1;
                int l = min(w, height[j]);
                maxArea = max(maxArea, l * l);
            }
        }
        return maxArea;
    }
    vector<vector<int>> getLessThan(vector<int>& nums){
        vector<vector<int>> res(nums.size(), vector<int>(2, -1));
        stack<int> s1;
        for(int i=0;i<nums.size();++i){
            int r = i;
            while(!s1.empty() && nums[s1.top()] >= nums[i]){
                int index = s1.top();
                s1.pop();
                int l = s1.empty() ? -1 : s1.top();
                res[index][0] = l;
                res[index][1] = r;
            }
            s1.push(i);
        }
        int r = nums.size();
        while(!s1.empty()){
            int index = s1.top();
            s1.pop();
            int l = s1.empty() ? -1 : s1.top();
            res[index][0] = l;
            res[index][1] = r;
        }
        return res;
    }
};

解法2:动态规划