Given an m x n
binary matrix
filled with 0
's and 1
's, find the largest square containing only 1
's and return its area.
Example 1:
Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]] Output: 4
Example 2:
Input: matrix = [["0","1"],["1","0"]] Output: 1
Example 3:
Input: matrix = [["0"]] Output: 0
Constraints:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 300
matrix[i][j]
is'0'
or'1'
.
最大正方形。
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。这是一道矩阵类的DP问题。dp[i][j]的定义是以坐标[i][j]为右下角的点组成的正方形的最大边长。状态转移方程是dp(i, j) = min(dp(i−1, j), dp(i−1, j−1), dp(i, j−1)) + 1。这样的题目练多了才会有思路。下图帮助理解DP矩阵是怎么被完成的。
时间O(n^2)
空间O(n^2)
Java一维实现
1 class Solution { 2 public int maximalSquare(char[][] matrix) { 3 int m = matrix.length; 4 int n = m > 0 ? matrix[0].length : 0; 5 int[] dp = new int[n + 1]; 6 int max = 0; 7 int prev = 0; 8 for (int i = 1; i <= m; i++) { 9 for (int j = 1; j <= n; j++) { 10 int temp = dp[j]; 11 if (matrix[i - 1][j - 1] == '1') { 12 dp[j] = Math.min(Math.min(dp[j - 1], prev), dp[j]) + 1; 13 max = Math.max(max, dp[j]); 14 } else { 15 dp[j] = 0; 16 } 17 prev = temp; 18 } 19 } 20 return max * max; 21 } 22 }
Java二维实现
1 class Solution { 2 public int maximalSquare(char[][] matrix) { 3 if (matrix.length == 0) return 0; 4 int m = matrix.length; 5 int n = matrix[0].length; 6 int res = 0; 7 int[][] dp = new int[m + 1][n + 1]; 8 for (int i = 1; i <= m; i++) { 9 for (int j = 1; j <= n; j++) { 10 if (matrix[i - 1][j - 1] == '1') { 11 dp[i][j] = Math.min(Math.min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]) + 1; 12 res = Math.max(res, dp[i][j]); 13 } 14 } 15 } 16 return res * res; 17 } 18 }
相关题目
84. Largest Rectangle in Histogram