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  • 【LeetCode】221. Maximal Square

    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.

    Credits:
    Special thanks to @Freezen for adding this problem and creating all test cases.

    这题的DP思想部分借鉴了jianchao.li.fighter

    思路如下:构建二维数组len,len[i][j]表示以(i,j)为右下角的最大方块的边长。

    递推关系为 len[i][j] = min(min(len[i-1][j], len[i][j-1]), len[i-1][j-1]) + 1;

    如下图示意:

    以(i,j)为右下角的最大方块边长,取决于周围三个位置(i-1,j),(i,j-1),(i-1,j-1),恰好为三者最小边长扩展1位。

    若三者最小边长为0,那么(i,j)自成边长为1的方块。

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

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  • 原文地址:https://www.cnblogs.com/ganganloveu/p/4632497.html
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