For example, given the following matrix:
1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
Return 4.
[分析]
思路1:修改MaxRectangle的实现,求解每一行上最大矩形面积改为求解最大正方形面积。
思路2:dp[i][j]表示以(i, j)为右下角顶点的正方形的边长,则(i, j)处为1时dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1
public class Solution { // Method 2: http://blog.csdn.net/xudli/article/details/46371673 public int maximalSquare(char[][] matrix) { if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return 0; int rows = matrix.length, cols = matrix[0].length; int max = 0; int[][] dp = new int[rows][cols]; for (int i = 0; i < rows; i++) { dp[i][0] = matrix[i][0] == '1' ? 1 : 0; max = Math.max(max, dp[i][0]); } for (int j = 0; j < cols; j++) { dp[0][j] = matrix[0][j] == '1' ? 1 : 0; max = Math.max(max, dp[0][j]); } for (int i = 1; i < rows; i++) { for (int j = 1; j < cols; j++) { if (matrix[i][j] == '1') { dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i][j - 1], dp[i - 1][j])) + 1; max = Math.max(max, dp[i][j]); } } } return max * max; } // Method 1: extension of maximalRectangle public int maximalSquare1(char[][] matrix) { if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return 0; int rows = matrix.length, cols = matrix[0].length; int[] h = new int[cols + 1]; int max = 0; for (int i = 0; i < rows; i++) { for (int j = 0; j < cols; j++) { if (matrix[i][j] == '1') h[j] += 1; else h[j] = 0; } max = Math.max(max, getMaxSquare(h)); } return max; } public int getMaxSquare(int[] h) { LinkedList<Integer> stack = new LinkedList<Integer>(); int max = 0; for (int i = 0; i < h.length; i++) { while (!stack.isEmpty() && h[i] < h[stack.peek()]) { int currH = h[stack.pop()]; while (!stack.isEmpty() && h[stack.peek()] == currH) stack.pop(); int len = Math.min(i - (stack.isEmpty() ? -1 : stack.peek()) - 1, currH); max = Math.max(max, len * len); } stack.push(i); } return max; } }