MaxArea-最大矩阵

MaxArea-最大矩阵

几道最大区域（矩形）相关的题。
题目一：

container-with-most-water
Given n non-negative integers a1 , a2 , …, an , where each represents a point at coordinate (i, ai ). n vertical lines are drawn such that the two endpoints of line i is at (i, ai ) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
Note: You may not slant the container.

题目二：
比较常见经典的一道题，求直方图的最大矩形

题目三：

maximal-rectangle
Given a 2D binary matrix filled with 0’s and 1’s, find the largest rectangle containing all ones and return its area.
例如二维数组矩阵：
[[1, 1, 0, 0, 1],
[0, 1, 0, 0, 1],
[0, 0, 1, 1, 1],
[0, 0, 1, 1, 1],
[0, 0, 0, 0, 1]]
输出6

下面直接上代码，代码可能更直观

import java.util.*;

public class MaxArea {


    //两边和x轴构成的container最大
    //height[i]之间宽为1
    //初始low=0，high=height.length-1，不断淘汰短的一边(以该边为边的最大container为当前area)，宽度不断减小
    //height[low]>height[high]时，比当前大的area，双边肯定在 low~high-1 之间选择
    //O(n)
    public static int maxArea(int[] height) {
        int low=0,high=height.length-1;
        int area =0,max=0;
        while(low<high){
            area=Math.max(area,(high-low)*Math.min(height[low],height[high]));
            if(height[low]>height[high])
                high--;
            else if(height[low]<height[high])
                low++;
            else{
                low++;
                high--;
            }
        }
        return area;
    }


    //直方图最大矩形，O(n)
    //主要思路：实际是中心扩展法的优化
    // j<i，heights[j]遇到更小的heights[i]时，可以求以j，heights[j]为高的最大矩形（以heights[j]为限制高，向两侧扩展）
    public static int largestRectangleArea(int[] heights) {
        if(heights==null||heights.length==0)
            return 0;
        int res=0;
        Stack<Integer> stack=new Stack<>();
        for(int i=0;i<heights.length;i++){
            while(!stack.isEmpty()&&heights[i]<=heights[stack.peek()]){
                int cur=stack.pop();
                int left=stack.isEmpty()?-1:stack.peek();
                res=Math.max(res,(i-left-1)*heights[cur]);
            }
            stack.push(i);
        }
        while(!stack.isEmpty()){
            int cur=stack.pop();
            int left=stack.isEmpty()?-1:stack.peek();
            res=Math.max(res,(heights.length-left-1)*heights[cur]);
        }
        return res;
    }

    //O(n*n)
    //添加转换：
    //    0 0 1 1 0 -> 0 0 1 1 0
    //    0 0 1 1 0 -> 0 0 2 2 0
    //    1 1 0 0 0 -> 1 1 0 0 0
    //    1 1 1 0 0 -> 2 2 1 0 0
    //再以每一行为底去求最大矩形
    public static int maximalRectangle(char[][] matrix) {

        if (matrix == null || matrix[0].length == 0) {
            return 0;
        }

        int[][] dp=new int[matrix.length][matrix[0].length];
        for(int i=0;i<matrix.length;i++){
            for(int j=0;j<matrix[0].length;j++)
            if(i==0){
                dp[i][j]=matrix[i][j]=='1'?1:0;
            }else{
                dp[i][j]=matrix[i][j]=='0'?0:dp[i-1][j]+1;
            }
        }

        int max=0;
        for(int i=0;i<matrix.length;i++){
            int temp=largestRectangleArea(dp[i]);
            max=Math.max(max,temp);
        }
        return max;

    }

}