标签：style mat solution col for fill res == put
Given a 2D binary matrix filled with 0‘s and 1‘s, find the largest rectangle containing only 1‘s and return its area.
Example:
Input: [ ["1","0","1","0","0"], ["1","0","1","1","1"], ["1","1","1","1","1"], ["1","0","0","1","0"] ] Output: 6
class Solution { public int maximalRectangle(char[][] matrix) { int le = matrix.length; int res = 0; if(le == 0) return 0; ArrayList<Integer> li = new ArrayList(); for(int i = 0; i < matrix[0].length; i++) li.add(0); for(int i = 0; i < le; i++){ for(int j = 0; j < matrix[0].length; j++){ if(matrix[i][j] == ‘0‘) li.set(j, 0); else li.set(j, li.get(j)+1); } int tmp = maxhisto(li); res = Math.max(tmp, res); } return res; } public int maxhisto(ArrayList<Integer> list){ int res = 0; int s = list.size(); for(int i = 0; i < s; i++){ if(i+1 < s && list.get(i+1)>=list.get(i)) continue; int mh = list.get(i); for(int j = i; j >= 0; j--){ mh = Math.min(mh, list.get(j)); int area = mh * (i - j + 1); res = Math.max(area, res); } } return res; } }
把每一列看成histogram，就可以得到4个histogram，从第一行开始，0，01，012，0123.把数值存入arraylist，如果是1就在原基础上+1，如果是0就置0.
然后用求maxhistogram里的方法，最简单的方法就是找局部顶点（数值大于右边）然后从局部顶点向左计算area面积。
标签：style mat solution col for fill res == put
原文地址：https://www.cnblogs.com/wentiliangkaihua/p/11645722.html
