leetcode 85 Maximal Rectangle ----- java

Posted 2020-08-15 xiaoba1203

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.

 

求矩阵中由1组成的最大子矩阵面积。

 

1、首先进行观察，发现如果直接遍历然后求出每一个点的最大矩阵明显是不对的，时间复杂度过高。

2、结合84题，发现这两道题十分类似，就题目中给出的例子可以仔细观察可以发现，最大子矩阵是以某一行作为底，然后用84题的方法，进行求解，所以先对这个矩阵进行转换，例子中的矩阵可以转换如下：

1 0 1 0 0　　　　
1 0 1 1 1
1 1 1 1 1　　　　
1 0 0 1 0

转换之后：

1 0 1 0 0
2 0 2 1 1
3 1 3 2 2
4 0 0 3 0

然后对每一行进行最大矩阵面积求解。

public class Solution {
    public int maximalRectangle(char[][] matrix) {
        int len1 = matrix.length;
        if( len1 == 0)
            return 0;
        int len2 = matrix[0].length;
        if( len2 == 0)
            return 0;
        int result = 0;
        int[][] heights = new int[len1][len2];
        for( int i = 0;i<len2;i++){
            if( matrix[0][i] == ‘1‘)
                heights[0][i] = 1;
        }
        for( int i = 1;i < len1;i++){
            for( int j = 0;j<len2;j++){
                if( matrix[i][j] == ‘1‘)
                    heights[i][j] = heights[i-1][j]+1;
            }
        }
        for( int i = 0;i<len1;i++){
            result = Math.max(result,largestRectangleArea(heights[i]));
        }
        return result;



    }
    public int largestRectangleArea(int[] heights) {


        int len = heights.length;
        int result = 0;
        if( len == 0)
            return 0;

        int[] left = new int[len];
        int[] right = new int[len];

        left[0] = 0;
        for( int i = 1;i<len;i++){
            int CurLeft = i-1;
            while( CurLeft >= 0 && heights[CurLeft]>=heights[i]){
                CurLeft = left[CurLeft]-1;
            }
            left[i] = CurLeft+1;
        }
        right[len-1] = len-1;
        for( int i = len-2;i>=0;i--){
            int CurRight = i+1;
            while( CurRight<len && heights[CurRight]>=heights[i])
                CurRight = right[CurRight]+1;
            right[i] = CurRight-1;
        }
        for( int i = 0;i<len;i++){
            result = Math.max(result,(right[i]-left[i]+1)*(heights[i]));
        }
        return result;}
}