文章目錄

• 題目描述
• 思路 && 代碼
• • • 二刷

題目描述

• 其實是84. 柱狀圖中最大的矩形的兄弟題目，理解成多個84題，對結果取max即可。
  
  

思路 && 代碼

• 一行抽象出一個【柱狀圖】，分別套到84題的函數里即可
• 時空復雜度：O($n^2$)、O(n)

class Solution {public int maximalRectangle(char[][] matrix) {if(matrix == null || matrix.length == 0) {return 0;}// 看成【對每層，進行柱狀圖最大面積判斷】即可（相當于固定底）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. 求柱狀圖最大矩陣面積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

• 思路還是記得的

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;// 逐行轉換for(int i = 0; i < matrix.length; i++) {// 當前行的逐列維護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;} }

總結

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