leetCode练习（85）

题目：Maximal Rectangle

难度：hard

问题描述：

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.

解题思路：
求最大全1矩形。可以化简为84题的子问题。参考第84题。以上图为例，以此计算0-0层，0-1层，0-2层····0-4层的最大矩阵即可。

具体代码如下：

public class h_85_MaximalRectangle {
public static int maximalRectangle(char[][] matrix) {
int[] temp=new int[matrix[0].length];
int res=0;
int temp2;
if(matrix.length==0||matrix[0].length==0){
return 0;
}
for(int i=0;i<temp.length;i++){
temp[i]=0;
}
for(int i=0;i<matrix.length;i++){
for(int j=0;j<temp.length;j++){
if(matrix[i][j]=='0'){
temp[j]=0;
}else{
temp[j]=1+temp[j];
}
}
for(int w:temp){
System.out.print(w+" ");
}
System.out.println();
temp2=largestRectangleArea(temp);
System.out.println(i+":"+temp2);
res=res>temp2?res:temp2;
}
return res;
}
public static int largestRectangleArea(int[] heights) {
int len=heights.length;
int res=0;
int temp;
int left=0,right=0;
int i;
Stack<Integer> stack=new Stack<>();
for(i=0;i<len;i++){
if(stack.isEmpty()){
stack.push(i);
continue;
}
while(!stack.isEmpty()){
if(heights[stack.peek()]<=heights[i]){
break;
}else{
temp=stack.pop();
right=i-1;
if(stack.isEmpty()){
left=0;
}else{
left=stack.peek()+1;
}
res=res>(right-left+1)*heights[temp]?res:(right-left+1)*heights[temp];
}
}
stack.push(i);
}
while(!stack.isEmpty()){
int top=heights[stack.pop()];
int size=top*(stack.isEmpty()?i:i-stack.peek()-1);
res=Math.max(res, size);
}
return res;
}
public static void main(String[]args){
char[][] a=new char[4][5];
char[]nums1={'1','0','1','0','0'};
char[]nums2={'1','0','1','1','1'};
char[]nums3={'1','1','1','1','1'};
char[]nums4={'1','0','0','1','0'};
a[0]=nums1;
a[1]=nums2;
a[2]=nums3;
a[3]=nums4;
maximalRectangle(a);
}
}