Leetcode 85. Maximal Rectangle 最大矩形 解题报告

1 解题思想

这道题我是转化成上一道题来做的,对于每一行,看成给一个直方图

在每一行上,首先按照直方图的方式进行统计,然后记录并更新最大,因此计算请看:
Leetcode 84. Largest Rectangle in Histogram 最大矩形 解题报告

2 原题

Given a 2D binary matrix filled with 0’s and 1’s, find the largest rectangle containing all ones and return its area.

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3 AC解

public class Solution {
      public int largestRectangleArea(int[] heights) {
        Stack<Integer> stack = new Stack<Integer>();
        int max_area=0,currentHeight,lastIndex,len;
        for(int i=0;i<heights.length;i++){
            currentHeight=heights[i];
            //保持升序
            if(stack.isEmpty() || heights[stack.peek()]<=currentHeight){
                stack.push(i);
                continue;
            }
            //到了这里,就是不满足升序,需要弹出,注意当前len的判断方式
            while(stack.isEmpty()==false && heights[stack.peek()]>currentHeight){
                lastIndex=stack.pop();
                len=stack.isEmpty()?i:(i-stack.peek()-1);
                max_area=Math.max(max_area,len*heights[lastIndex]);
            }
            stack.push(i);
        }
        //最后需要多处理一次
        while(stack.isEmpty()==false){
            lastIndex=stack.pop();
            len=stack.isEmpty()?heights.length:(heights.length-stack.peek()-1);
            max_area=Math.max(max_area,len*heights[lastIndex]);
        }
        return max_area;

    }
    //将问题变为条上一题
    public int maximalRectangle(char[][] matrix) {
        int n=matrix.length;
        int max=0;
        if(n<1) return 0;
        int m=matrix[0].length;
        if(m<1) return 0;
        for(int i=0;i<n;i++){
            int heights[]=new int[m];
            for(int j=0;j<m;j++){
                int t=i;
                while(t>=0 && matrix[t--][j]=='1') heights[j]++;
            }
            max=Math.max(largestRectangleArea(heights),max);
        }
        return max;
    }
}

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