leecode 85 最大矩形 hard
作者:互联网
给定一个仅包含 0 和 1 、大小为 rows x cols 的二维二进制矩阵,找出只包含 1 的最大矩形,并返回其面积。
示例 1:
输入:matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
输出:6
解释:最大矩形如上图所示。
示例 2:
输入:matrix = []
输出:0
示例 3:
输入:matrix = [["0"]]
输出:0
示例 4:
输入:matrix = [["1"]]
输出:1
示例 5:
输入:matrix = [["0","0"]]
输出:0
提示:
rows == matrix.length
cols == matrix[0].length
0 <= row, cols <= 200
matrix[i][j] 为 '0' 或 '1'
暴力解法
public int maximalRectangle(char[][] matrix)
{
if(matrix.length==0) return 0;
int maxArea=0;
int[][] new_maxtrix= new int[matrix.length][matrix[0].length];
for(int i=0;i<new_maxtrix.length;i++)
{
for(int j=0;j<new_maxtrix[0].length;j++)
{
if(matrix[i][j]=='1')
{
if(j==0)
{
new_maxtrix[i][j]=1;
}else {
new_maxtrix[i][j] = new_maxtrix[i][j - 1] + 1;
}
}else
{
new_maxtrix[i][j]=0;
}
int minwidth=new_maxtrix[i][j];
int height;
for(int k=i;k>=0;k--)
{
height=i-k+1;
minwidth=Math.min(minwidth,new_maxtrix[k][j]);
maxArea=Math.max(maxArea,minwidth*height);
}
}
}
return maxArea;
}
单调栈
解题思路
https://www.cnblogs.com/AI-Creator/p/14767387.html
public int compute(int[][] arr,int row)
{
int ans=0;
LinkedList<Integer> stack= new LinkedList<>();
//每一行单调递增栈
int l,r;
for(int i=0;i<arr[0].length;i++)
{
while (!stack.isEmpty()&&arr[row][i]<arr[row][stack.peek()])
{
int curr=stack.pop();
l=stack.peek();
r=i;
ans= Math.max(ans,(r-l-1)*arr[row][curr]);
}
stack.push(i);
}
return ans;
}
public int maximalRectangle(char[][] matrix) {
if(matrix.length==0) return 0;
int[][] height_matrix = new int[matrix.length][matrix[0].length+2];
//构造高度矩阵
for(int i=1;i<height_matrix[0].length-1;i++)
{
height_matrix[0][i]=matrix[0][i-1]-'0';
}
for(int i=1;i<height_matrix.length;i++)
{
for(int j=1;j<height_matrix[0].length-1;j++)
{
if(matrix[i][j-1]=='0')continue;
height_matrix[i][j]=height_matrix[i-1][j]+1;
}
}
int max= 0;
for(int i=0;i<height_matrix.length;i++)
{
max=Math.max(compute(height_matrix,i),max);
int a=5;
}
return max;
}
动态规划?
标签:matrix,示例,int,hard,length,leecode,new,maxArea,85 来源: https://www.cnblogs.com/AI-Creator/p/14772102.html