LeetCode221 最大正方形

题目

在一个由 '0' 和 '1' 组成的二维矩阵内，找到只包含 '1' 的最大正方形，并返回其面积。

 示例 1： 
输入：matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"]
,["1","0","0","1","0"]]
输出：4
 示例 2： 
输入：matrix = [["0","1"],["1","0"]]
输出：1
 示例 3： 
输入：matrix = [["0"]]
输出：0

 提示： 
 m == matrix.length 
 n == matrix[i].length 
 1 <= m, n <= 300 
 matrix[i][j] 为 '0' 或 '1' 

方法

动态规划法

dp[i][j]表述以坐标（i，j）为右下角的最大边长
坐标（i，j）的值取决于左边、上面、左上角的最小值
dp[i][j] = Math.min(Math.min(dp[i-1][j],dp[i][j-1]),dp[i-1][j-1])+1;

• 时间复杂度：O(mn)，m为行数，n为列数
• 空间复杂度：O(mn)

class Solution {
    public int maximalSquare(char[][] matrix) {
        int row = matrix.length,col = matrix[0].length;
        int[][] dp = new int[row][col];
        int maxSide = 0;
        for(int i=0;i<row;i++){
            for(int j = 0;j<col;j++){
                if(i==0||j==0){
                    dp[i][j] = matrix[i][j]-'0';
                }else if(matrix[i][j]=='1'){
                    dp[i][j] = Math.min(Math.min(dp[i-1][j],dp[i][j-1]),dp[i-1][j-1])+1;
                }
                maxSide = Math.max(maxSide,dp[i][j]);
            }
        }
        return maxSide*maxSide;
    }
}