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Q221MaximalSquare.java
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Q221MaximalSquare.java
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/**
* @author ahscuml
* @date 2019/1/17
* @time 15:50
*/
public class Q221MaximalSquare {
/**
* 主函数
*/
public static void main(String[] args) {
char[][] matrix = {{'1', '0', '1', '0', '0'}, {'1', '0', '1', '1', '1'}, {'1', '1', '1', '1', '1'}, {'1',
'0', '0', '1', '0'}};
System.out.println("最大的面积是" +
";" + maximalSquare(matrix));
System.out.println("最大的面积是" + maximalSquareII(matrix));
}
/**
* 利用动态规划的方法
* 重点是要找到equation :we define the state as the maximal size of the square that can be achieved at point (i, j)
* 我自己的想法,写起来很啰嗦
*/
public static int maximalSquare(char[][] matrix) {
if (matrix.length == 0) {
return 0;
}
int m = matrix.length;
int n = matrix[0].length;
// 创建新的辅助数组,用于存储当前的元素所能组成的最大的矩形边长
int dp[][] = new int[m][n];
int res = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
// 边界的直接赋原值
if (i == 0 || j == 0) {
if (matrix[i][j] == '0') {
dp[i][j] = 0;
} else {
dp[i][j] = 1;
}
} else {
// 如果当前的是1
if (matrix[i][j] == '1') {
// 如果左上角还是1
if (matrix[i - 1][j - 1] == '1') {
dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1], dp[i - 1][j]), dp[i][j - 1]) + 1;
} else {
// 左上角不是1,那当前就是1
dp[i][j] = 1;
}
// 如果当前是0,那就是0了
} else {
dp[i][j] = 0;
}
}
res = Math.max(res, dp[i][j]);
}
}
return res * res;
}
/**
* 更加快速简便的方法,思路都是相同的,通过增加长和宽来完成减小复杂度
*/
public static int maximalSquareII(char[][] matrix) {
if (matrix.length == 0) return 0;
int rows = matrix.length, cols = matrix[0].length;
int[][] dp = new int[rows + 1][cols + 1];
int maxsqlen = 0;
// dp[i][j]对应着matrix[i - 1][j - 1]
// 新的数组默认值是0
for (int i = 1; i <= rows; i++) {
for (int j = 1; j <= cols; j++) {
if (matrix[i - 1][j - 1] == '1') {
dp[i][j] = Math.min(Math.min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
maxsqlen = Math.max(maxsqlen, dp[i][j]);
}
}
}
return maxsqlen * maxsqlen;
}
}