yasinshaw
4247e0700d
Changes: - Removed all Java code implementations - Kept only Go language solutions - Renamed "## Go 解法" to "## 解法" - Removed "### Go 代码要点" sections - Cleaned up duplicate headers and empty sections - Streamlined documentation for better readability Updated files (9): - 三数之和.md - 两数相加.md - 无重复字符的最长子串.md - 最长回文子串.md - 括号生成.md - 子集.md - 单词搜索.md - 电话号码的字母组合.md - 柱状图中最大的矩形.md All 22 LeetCode Hot 100 Medium problems now use Go exclusively. Code is cleaner, more focused, and easier to follow. Generated with [Claude Code](https://claude.ai/code) via [Happy](https://happy.engineering) Co-Authored-By: Claude <noreply@anthropic.com> Co-Authored-By: Happy <yesreply@happy.engineering>
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柱状图中最大的矩形 (Largest Rectangle in Histogram)
题目描述
给定 n 个非负整数，用来表示柱状图中各个柱子的高度。每个柱子彼此相邻，且宽度为 1。
求在该柱状图中，能够勾勒出来的矩形的最大面积。
解题思路
方法一：单调栈（推荐）
**核心思想：**使用单调递增栈，存储柱子的索引。当遇到比栈顶小的柱子时，弹出栈顶并计算面积。
代码实现
Go 实现
package main
func largestRectangleArea(heights []int) int {
stack := []int{}
maxArea := 0
n := len(heights)
for i := 0; i <= n; i++ {
h := 0
if i < n {
h = heights[i]
}
for len(stack) > 0 && h < heights[stack[len(stack)-1]] {
height := heights[stack[len(stack)-1]]
stack = stack[:len(stack)-1]
width := i
if len(stack) > 0 {
width = i - stack[len(stack)-1] - 1
}
area := height * width
if area > maxArea {
maxArea = area
}
}
stack = append(stack, i)
}
return maxArea
}
- LeetCode 85: 最大矩形（二维版本）
- LeetCode 42: 接雨水