Largest Rectangle in Histogram leetcode java

最新推荐文章于 2025-09-05 16:34:31 发布

转载最新推荐文章于 2025-09-05 16:34:31 发布 · 78 阅读

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#java #数据结构与算法

本文介绍了一种高效算法来解决寻找直方图中最大矩形面积的问题，并提供了两种解决方案，一种为O(n^2)，另一种为更优的O(n)算法实现，后者巧妙地运用了栈的数据结构。

题目：

Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].

The largest rectangle is shown in the shaded area, which has area = 10 unit.

For example,
Given height = [2,1,5,6,2,3],
return 10.

题解：

这道题自己是完全没想到用栈了。。

有个很完整很详细很好的讲解在这里： http://www.cnblogs.com/lichen782/p/leetcode_Largest_Rectangle_in_Histogram.html

我就不写了。贴一下上面提到的代码吧。

O(n^2)的：

1 public int largestRectangleArea(int[] height) {
2         // Start typing your Java solution below
3         // DO NOT write main() function
4         int[] min = new int[height.length];
5         int maxArea = 0;
6         for(int i = 0; i < height.length; i++){
7             if(height[i] != 0 && maxArea/height[i] >= (height.length - i)) {
8                 continue;
9             }
10             for(int j = i; j < height.length; j++){
11                 if(i == j) min[j] = height[j];
12                 else {
13                     if(height[j] < min[j - 1]) {
14                         min[j] = height[j];
15                     }else min[j] = min[j-1];
16                 }
17                 int tentativeArea = min[j] * (j - i + 1);
18                 if(tentativeArea > maxArea) {
19                     maxArea = tentativeArea;
20                 }
21             }
22         }
23         return maxArea;
24     }

O(n)的：

1 public int largestRectangleArea2(int[] height) {
2         Stack<Integer> stack = new Stack<Integer>();
3         int i = 0;
4         int maxArea = 0;
5         int[] h = new int[height.length + 1];
6         h = Arrays.copyOf(height, height.length + 1);
7         while(i < h.length){
8             if(stack.isEmpty() || h[stack.peek()] <= h[i]){
9                 stack.push(i++);
10             }else {
11                 int t = stack.pop();
12                 maxArea = Math.max(maxArea, h[t] * (stack.isEmpty() ? i : i - stack.peek() - 1));
13             }
14         }
15         return maxArea;
16     }