Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area.
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 6.
題意：找到一個(gè)矩陣中卢未，值全部是1的最大矩形橡淆。
思路：這道題首先是肯定可以暴力求解的瘩扼，以矩陣中每個(gè)是1元素為矩形的左上角，邊長從2到n，往它的右下找出最大的矩形暇番，但是這種時(shí)間復(fù)雜度應(yīng)該是n的四次方。
看了discuss，思路是對每一行锈津，把它上面的所有行往下做投影，比如上面的矩形會(huì)變成：
10100
20211
31322
40030
然后對每一行用84題Largest Rectangle in Histogram的方法求解最大矩陣凉蜂。
public int maximalRectangle(boolean[][] matrix) {
// Write your code here
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return 0;
}
int n = matrix.length;
int m = matrix[0].length;
int[][] height = new int[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (i == 0) {
height[i][j] = matrix[i][j] ? 1 : 0;
} else {
height[i][j] = matrix[i][j] ? 1 + height[i-1][j] : 0;
}
}
}
int res = 0;
for (int i = 0; i < n; i++) {
Stack<Integer> stack = new Stack<Integer>();
for (int j = 0; j <= m; j++) {
int nowHeight = j == m ? -1 : height[i][j];
while (!stack.isEmpty() && height[i][stack.peek()] >= nowHeight) {
int h = height[i][stack.pop()];
int w = stack.isEmpty() ? j : j - stack.peek() - 1;
res = Math.max(res, h*w);
}
stack.push(j);
}
}
return res;
}