MaximalRectangle

1.问题描述

给定一个由'0'和'1'组成的2D空间，找到这个空间中只包含1的最大矩阵

2.解题思路

将此问题转换为求只包含'1'的左边界和右边界问题

3.程序源码

public int maximalRectangle(char[][] matrix) {
if(matrix==null || matrix.length==0 || matrix[0].length==0)
return 0;
int max = 0;
int rows = matrix.length; //行数
int cols = matrix[0].length;//列数
int[] height = new int[cols];//第i行第j列直方图的高
int[] left = new int[cols]; //左边界(与高度相关进行判断)
int[] right = new int[cols]; //右边界
for(int i=0; i<cols; i++){
right[i] = cols;
}
for(int i=0; i<rows; i++){//第i层
int curLeft = 0;
int curRight = cols; //当前左边界和右边界
//更新左边界
for(int j=0; j<cols; j++){
if(matrix[i][j] == '1'){
height[j]++;
left[j] = Math.max(curLeft, left[j]);
}else{
height[j] = 0;
curLeft = j+1;
left[j] = 0;
}
}
for(int j=cols-1; j>=0; j--){ //更新右边界
if(matrix[i][j]=='1'){
right[j] = Math.min(curRight, right[j]); //很重要的一步
}else{
right[j] = cols;
curRight = j;
}
//max = Math.max(max, (right[j]-left[j])*height[j]);
}
for(int k=0;k<cols;k++){
if(max<(right[k]-left[k])*height[k]){
max=(right[k]-left[k])*height[k];
}
}
}
return max;
}

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