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.

DP

Time Complexity
O(n^n)
Space Complexity
O(n^n)

思路

size 1
1
size 2
1 1
1 1
size 3
1 1 1
1 1 1
1 1 1

dpi the largest all 1 square's size withi as the bottom right (including)/up left
base case
dp0 dp0 dpi
dp rule
dpi = min(dpi-1, dpi, dpi - 1 if matrixi == 1 else 0
keep one global_max

代码

public int maximalSquare(char[][] matrix) {
    //corner case
    if(matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0) return 0;
    int rows = matrix.length, cols = matrix[0].length;
    int[][] dp = new int[rows][cols];
    int max = 0;
    
    for(int i = 0; i < rows; i++){
        if(matrix[i][0] == '1') dp[i][0] = 1;
        max = Math.max(dp[i][0], max);
    }
    for(int j = 1; j < cols; j++){
        if(matrix[0][j] == '1') dp[0][j] = 1;
        max = Math.max(dp[0][j], max);
    }

    for(int i = 1; i < rows; i++){
        for(int j = 1; j < cols; j++){
                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;
                    max = Math.max(dp[i][j], max);
                }
        }
    }
    return max*max;
}

Test Cases

null
{}
{null, ...}
{{}, ...}
{0}
{1}
{{0 0 0 1 0}}
{{0}，{0}，{0}，{1}，{0}}