1 解題思想

這道題我是轉化成上一道題來做的，對於每一行，看成給一個直方圖

2 原題

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
 
;
            }
            //到了這裡，就是不滿足升序，需要彈出，注意當前len的判斷方式
            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);
        }
        //最後需要多處理一次
        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;

    }
    //將問題變為條上一題
    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;
    }
}