Loading

Medium | LeetCode 221. 最大正方形 | 动态规划

221. 最大正方形

在一个由 '0''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:

img
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:

img
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]'0''1'

动态规划

状态转移方程:

\[d p(i, j)=\min (d p(i-1, j), d p(i-1, j-1), d p(i, j-1))+1 \]

public int maximalSquare(char[][] matrix) {
    int maxSide = 0;
    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
        return maxSide;
    }
    int rows = matrix.length, columns = matrix[0].length;
    int[][] dp = new int[rows][columns];
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < columns; j++) {
            if (matrix[i][j] == '1') {
                if (i == 0 || j == 0) {
                    dp[i][j] = 1;
                } else {
                    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]);
            }
        }
    }
    int maxSquare = maxSide * maxSide;
    return maxSquare;
}
posted @ 2021-01-27 11:49  反身而诚、  阅读(45)  评论(0编辑  收藏  举报