LeetCode OJ 之 Maximal Square （最大的正方形）

题目：

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing all 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 4.

题目链接：https://leetcode.com/problems/maximal-square/。

这一题有点相似：LeetCode OJ 之 Maximal Rectangle （最大的矩形）。可是解题方法全然不同。

思路：

动态规划。设f[i][j]表示包含当前点的正方形的最大变长，有递推关系例如以下：

f[0][j] = matrix[0][j] 
f[i][0] = matrix[i][0] 
For i > 0 and j > 0: 
if matrix[i][j] = 0, f[i][j] = 0; 
if matrix[i][j] = 1, f[i][j] = min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1.

代码1：

class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) 
    {
        int row = matrix.size();
        if(row == 0)
            return 0;
        int col = matrix[0].size();
        vector<vector<int> > f(row , vector<int>(col , 0));
        int maxsize = 0;    //最大边长
        for(int i = 0 ; i < row ; i++)
        {
            for(int j = 0 ; j < col ; j++)
            {
                if(i == 0 || j == 0)
                    f[i][j] = matrix[i][j]-'0';
                else
                {
                    if(matrix[i][j] == '0')
                        f[i][j] = 0;
                    else
                        f[i][j] = min(min(f[i-1][j] , f[i][j-1]) , f[i-1][j-1]) + 1;
                }
                maxsize = max(maxsize , f[i][j]);
            }
        }
        return maxsize * maxsize;
    }
    
};

代码2：

优化空间为一维

class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) 
    {
        int row = matrix.size();
        if(row == 0)
            return 0;
        int col = matrix[0].size();
        
        vector<int> f(col , 0);
        
        int tmp1 = 0 , tmp2 = 0;
        
        int maxsize = 0;    //最大边长
        for(int i = 0 ; i < row ; i++)
        {
            for(int j = 0 ; j < col ; j++)
            {
                tmp1 = f[j];    //tmp1把当前f[j]保存以下，用来做下一次推断f[i+1][j+1]的左上角f[i-1][j-1]
                if(i == 0 || j == 0)
                    f[j] = matrix[i][j]-'0';
                else
                {
                    if(matrix[i][j] == '0')
                        f[j] = 0;
                    else
                        f[j] = min(min(f[j-1] , f[j]) , tmp2) + 1;  //这里的tmp2即是代码1的f[i-1][j-1]
                }
                tmp2 = tmp1 ;   //把tmp1赋给tmp2，用来下次for循环求f[j+1]
                maxsize = max(maxsize , f[j]);
            }
        }
        return maxsize * maxsize;
    }
    
};