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修改 0084.柱状图中最大的矩形 Go单调栈代码的格式

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ShuangmingMa
2023-09-16 22:33:02 +08:00
committed by GitHub
parent 8c3f064927
commit c26fddc702
1 changed files with 27 additions and 29 deletions
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56
problems/0084.柱状图中最大的矩形.md
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@ -478,36 +478,34 @@ class Solution:
```go
func largestRectangleArea(heights []int) int {
// 声明max并初始化为0
max := 0
// 使用切片实现栈
stack := make([]int, 0)
// 数组头部加入0
heights = append([]int{0}, heights...)
// 数组尾部加入0
heights = append(heights, 0)
// 初始化栈，序号从0开始
stack = append(stack, 0)
for i := 1; i < len(heights); i++ {
// 结束循环条件为：当即将入栈元素>top元素，也就是形成非单调递增的趋势
for heights[stack[len(stack)-1]] > heights[i] {
// mid 是top
mid := stack[len(stack)-1]
// 出栈
stack = stack[0 : len(stack)-1]
// left是top的下一位元素，i是将要入栈的元素
left := stack[len(stack)-1]
// 高度x宽度
tmp := heights[mid] * (i - left - 1)
if tmp > max {
max = tmp
}
}
stack = append(stack, i)
}
return max
max := 0
// 使用切片实现栈
stack := make([]int, 0)
// 数组头部加入0
heights = append([]int{0}, heights...)
// 数组尾部加入0
heights = append(heights, 0)
// 初始化栈，序号从0开始
stack = append(stack, 0)
for i := 1; i < len(heights); i ++ {
// 结束循环条件为：当即将入栈元素>top元素，也就是形成非单调递增的趋势
for heights[stack[len(stack) - 1]] > heights[i] {
// mid 是top
mid := stack[len(stack) - 1]
// 出栈
stack = stack[0 : len(stack) - 1]
// left是top的下一位元素，i是将要入栈的元素
left := stack[len(stack) - 1]
// 高度x宽度
tmp := heights[mid] * (i - left - 1)
if tmp > max {
max = tmp
}
}
stack = append(stack, i)
}
return max
}
```
### JavaScript: