53 lines
2.2 KiB
Java
53 lines
2.2 KiB
Java
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package dynamic_programming;
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/**
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* 题目： 221. 最大正方形 (maximalSquare)
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* 描述：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。
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*
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示例 1：
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输入：matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
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输出：4
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* 链接：https://leetcode.cn/problems/maximal-square/
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*/
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//不会
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public class MaximalSquare {
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/**
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* 那么如何计算 dp 中的每个元素值呢？对于每个位置 (i,j)，检查在矩阵中该位置的值：
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*
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* 如果该位置的值是 0，则 dp(i,j)=0，因为当前位置不可能在由 1 组成的正方形中；
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*
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* 如果该位置的值是 1，则 dp(i,j) 的值由其上方、左方和左上方的三个相邻位置的 dp 值决定。具体而言，当前位置的元素值等于三个相邻位置的元素中的最小值加 1，状态转移方程如下：
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*
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* dp(i,j)=min(dp(i−1,j),dp(i−1,j−1),dp(i,j−1))+1
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*
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* 作者：力扣官方题解
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* 链接：https://leetcode.cn/problems/maximal-square/solutions/234964/zui-da-zheng-fang-xing-by-leetcode-solution/
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* 来源：力扣（LeetCode）
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* 著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
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* @param matrix
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* @return
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*/
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public int maximalSquare(char[][] matrix) {
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int maxSide = 0;
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if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
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return maxSide;
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}
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int rows = matrix.length, columns = matrix[0].length;
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int[][] dp = new int[rows][columns];
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for (int i = 0; i < rows; i++) {
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for (int j = 0; j < columns; j++) {
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if (matrix[i][j] == '1') {
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if (i == 0 || j == 0) {
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dp[i][j] = 1;
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} else {
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dp[i][j] = Math.min(Math.min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]) + 1;
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}
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maxSide = Math.max(maxSide, dp[i][j]);
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}
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}
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}
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int maxSquare = maxSide * maxSide;
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return maxSquare;
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}
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}
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