Leetcode221. Maximal Square

Leetcode221. 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.

Example:

Input: 

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Output: 4

Dynamic Programming

Can be considered a square of the number 3 is a square of 2 stacked obtained, which is the core of the state transition equation solving the problem.

First, the definition states:

With dp[i][j]expressed matrix[i][j]as the maximum length of the bottom right side of the square.

Second, the state transition equation:

Left, top, top left to take the smallest of the three, plus 1.

• Figure 1: is limited to the upper left of the 0
• Figure 2: is limited to the upper side 0
• Figure 3: is limited to the left of the 0
• Digital said: this is the largest side the lower right corner of the square long
• Yellow indicates:? Lower right corner of the grid as a square area

which is:dp[i][j] = Min(dp[i-1][j],dp[i][j-1],dp[i-1][j-1]) + 1

Third, the initialization and output

Initializing a first row and first column of dp[i][j] = matrix[i][j] - 0output d[i,j]. However, in order to avoid side determination process, the left most plus one f[i][0] = 0, plus one line on the left edge f[0][j] = 0.

public int maximalSquare(char[][] matrix) {
    int rows = matrix.length;
    if (rows == 0) {
        return 0;
    }
    int cols = matrix[0].length;
    int[][] dp = new int[rows + 1][cols + 1];
    int maxSide = 0;
    for (int i = 1; i <= rows; i++) {
        for (int j = 1; j <= cols; j++) {
            //因为多申请了一行一列，所以这里下标要减 1
            if (matrix[i - 1][j - 1] == '0') {
                dp[i][j] = 0;
            } else {
                dp[i][j] = Math.min(dp[i - 1][j], Math.min(dp[i][j - 1], dp[i - 1][j - 1])) + 1;
                maxSide = Math.max(dp[i][j], maxSide);
            }
        }
    }
    return maxSide * maxSide;
}

State compression

Because when updating the current row, only use the information of the previous row, before the row will not need again, so we can use one-dimensional array, no two-dimensional matrix.

public int maximalSquare(char[][] matrix) {
    int rows = matrix.length;
    if (rows == 0) {
        return 0;
    }
    int cols = matrix[0].length;
    int[] dp = new int[cols + 1];
    int maxSide = 0;
    int pre = 0;//保存左上角的值
    for (int i = 1; i <= rows; i++) {
        for (int j = 1; j <= cols; j++) {
            int temp = dp[j];
            if (matrix[i - 1][j - 1] == '0') {
                dp[j] = 0;
            } else {
                dp[j] = Math.min(dp[j - 1], Math.min(dp[j], pre)) + 1;
                maxSide = Math.max(dp[j], maxSide);
            }
            pre = temp;
        }
    }
    return maxSide * maxSide;
}