题目描述

• 其实是84. 柱状图中最大的矩形的兄弟题目，理解成多个84题，对结果取max即可。
  
  

思路 && 代码

• 一行抽象出一个【柱状图】，分别套到84题的函数里即可
• 时空复杂度：O($n^2$)、O(n)

class Solution {public int maximalRectangle(char[][] matrix) {if(matrix == null || matrix.length == 0) {return 0;}// 看成【对每层，进行柱状图最大面积判断】即可（相当于固定底）int max = 0;int[] heights = new int[matrix[0].length];for(int i = 0; i < matrix.length; i++) {for(int j = 0; j < matrix[0].length; j++) {if(matrix[i][j] == '1') {heights[j]++;}else {heights[j] = 0;}}max = Math.max(max, largestRectangleArea(heights));}return max;}// 84. 求柱状图最大矩阵面积public int largestRectangleArea(int[] heights) {int res = 0;Deque<Integer> stack = new ArrayDeque<>();int[] newHeights = new int[heights.length + 2];for(int i = 1; i < heights.length + 1; i++) {newHeights[i] = heights[i - 1];}for(int i = 0; i < newHeights.length; i++) {while(!stack.isEmpty() && newHeights[stack.peek()] > newHeights[i]) {int index = stack.pop();int l = stack.peek();int r = i;res = Math.max(res, (r - l - 1) * newHeights[index]);}stack.push(i);}return res;}
}

二刷

• 思路还是记得的

class Solution {public int maximalRectangle(char[][] matrix) {if(matrix == null || matrix.length == 0) return 0;int[] heights = new int[matrix[0].length];int res = 0;// 逐行转换for(int i = 0; i < matrix.length; i++) {// 当前行的逐列维护for(int j = 0; j < matrix[0].length; j++) {if(matrix[i][j] == '1') {heights[j]++;}else {heights[j] = 0;}}res = Math.max(res, largestRectangleArea(heights));}return res;}public int largestRectangleArea(int[] heights) {int[] newHeights = new int[heights.length + 2];for(int i = 1; i <= heights.length; i++) {newHeights[i] = heights[i - 1];}Deque<Integer> stack = new ArrayDeque<>();int max = 0;for(int i = 0; i < newHeights.length; i++) {while(!stack.isEmpty() && newHeights[i] < newHeights[stack.peek()]) {int now = stack.poll();int left = stack.peek(); int right = i; max = Math.max(max, (right - left - 1) * newHeights[now]);}stack.push(i);}return max;}
}