地址：https://oj.leetcode.com/problems/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.

首先吐槽下這個(gè)題目，不知道是我英語(yǔ)理解還是題目本身存在讓人誤解地方，我理解題意是最大的矩形包含全部1元素，按著這個(gè)理解我非常開心的寫了例如以下代碼：代碼思路非常easy，類似楊氏矩陣中用到的思想：http://blog.csdn.net/huruzun/article/details/28420545

public class Solution {private char[][] matrix ;public int maximalRectangle(char[][] matrix) {if(matrix.length == 0){return 0;}int row = matrix.length;int col = matrix[0].length;int ans = 0;this.matrix = matrix;Point start = new Point();Point end = new Point(row-1,col-1);while(true){if(isAllZeroDown(start, row)){start.y = start.y+1 >=col ? col-1:start.y+1;if(start.equals(end)){return 0;}}if(isAllZeroLeft(start, col)){start.x = start.x+1 >=row ? row-1:start.x+1;if(start.equals(end)){return 0;}}if(!isAllZeroDown(start, row) && !isAllZeroLeft(start, col)){break;}}while(true){if(isAllZeroRight(end)){end.y = end.y-1>= 0 ? end.y-1:0;}if(isAllZeroUp(end)){end.x = end.x-1>= 0 ? end.x-1:0;}if(!isAllZeroRight(end) && !isAllZeroUp(end)){break;}}ans = Math.abs((end.x-start.x+1)*(end.y - start.y+1));return ans;}boolean isAllZeroDown(Point p,int len){int j = p.y;for(int i=p.x;i<len;i++){if(matrix[i][j]=='1'){return false;}}return true;}boolean isAllZeroLeft(Point p,int len){int i = p.x;for(int j=p.y;j<len;j++){if(matrix[i][j]=='1'){return false;}}return true;}boolean isAllZeroUp(Point p){int i = p.x;for(int j=p.y;j>=0;j--){if(matrix[i][j]=='1'){return false;}}return true;}boolean isAllZeroRight(Point p){int j = p.y;for(int i=p.x;i>=0;i--){if(matrix[i][j]=='1'){return false;}}return true;} }

提交之后始終不正確，然后開始網(wǎng)上搜索，才發(fā)現(xiàn)題目正確意思是：

找一個(gè)最大矩陣，里面所有是1。

比例如以下圖：

依據(jù)這個(gè)圖轉(zhuǎn)換能夠發(fā)現(xiàn)問題，看下圖：

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轉(zhuǎn)到這個(gè)圖就會(huì)發(fā)現(xiàn)，這就是：http://blog.csdn.net/huruzun/article/details/39717501里面說(shuō)到的同樣問題。

java代碼：

public class Solution {public int largestRectangleArea(int[] height) {int maxarea = 0;Stack<Integer> sta = new Stack<>();int top ;int top_area;int i = 0;while(i<height.length){if(sta.isEmpty() || height[sta.peek()]<=height[i] ){sta.push(i++);}else{top = sta.pop();top_area = height[top] * (sta.isEmpty()? i:i-sta.peek()-1);if(top_area>maxarea){maxarea = top_area;}}}while(!sta.isEmpty()){top = sta.pop();top_area = height[top] * (sta.isEmpty()? i:i-sta.peek()-1);if(top_area>maxarea){maxarea = top_area;}}return maxarea;}public int maximalRectangle(char[][] matrix){if(matrix.length == 0){return 0;}int row = matrix.length;int col = matrix[0].length;int [][]dp = new int[row][];for(int i=0;i<row;i++){dp[i] = new int[col];}for(int j=0;j<col;j++){if(matrix[0][j]=='1'){dp[0][j] = 1;}}for(int j=0;j<col;j++){for(int i=1;i<row;i++){if(matrix[i][j] == '1'){dp[i][j] = dp[i-1][j]+1;}}}int maxarea = 0;for(int i=0;i<row;i++){int temp = largestRectangleArea(dp[i]);if(temp>maxarea){maxarea = temp;}}return maxarea;} }

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