[LeetCode] Maximal Square

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.

解题思路

1) Construct a sum matrix S[R][C] for the given M[R][C].     a) Copy first row and first columns as it is from M[][] to S[][]     b) For other entries, use following expressions to construct S[][]         If M[i][j] is 1 then            S[i][j] = min(S[i][j-1], S[i-1][j], S[i-1][j-1]) + 1         Else /*If M[i][j] is 0*/            S[i][j] = 02) Find the maximum entry in S[R][C]3) Using the value and coordinates of maximum entry in S[i], print    sub-matrix of M[][]

更详细的解题思路见GeeksforGeeks。

实现代码

class Solution {public:    int maximalSquare(vector<vector<char>>& matrix) {        if (matrix.empty())        {            return 0;        }        int row = matrix.size();        int col = matrix[0].size();        vector<vector<int>> s(row, vector<int>(col, 0));        for (int i = 0; i < row; i++)        {            if (matrix[i][0] == '1')            {                s[i][0] = 1;            }        }        for (int i = 0; i < col; i++)        {            if (matrix[0][i] == '1')            {                s[0][i] = 1;            }        }        for (int i = 1; i < row; i++)        {            for (int j = 1; j < col; j++)            {                if (matrix[i][j] == '1')                {                    s[i][j] = min(s[i-1][j], min(s[i][j-1], s[i-1][j-1])) + 1;                }                else                {                    s[i][j] = 0;                }            }        }        int width = 0;        for (int i = 0; i < row; i++)        {            for (int j = 0; j < col; j++)            {                width = max(width, s[i][j]);            }        }        return width * width;    }};