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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
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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 ```go
func largestRectangleArea(heights []int) int { func largestRectangleArea(heights []int) int {
// 声明max并初始化为0 max := 0
max := 0 // 使用切片实现栈
// 使用切片实现栈 stack := make([]int, 0)
stack := make([]int, 0) // 数组头部加入0
// 数组头部加入0 heights = append([]int{0}, heights...)
heights = append([]int{0}, heights...) // 数组尾部加入0
// 数组尾部加入0 heights = append(heights, 0)
heights = append(heights, 0) // 初始化栈，序号从0开始
// 初始化栈，序号从0开始 stack = append(stack, 0)
stack = append(stack, 0) for i := 1; i < len(heights); i ++ {
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
// 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] {
for heights[stack[len(stack)-1]] > heights[i] { // mid 是top
// mid 是top mid := stack[len(stack) - 1]
mid := stack[len(stack)-1] // 出栈
// 出栈 stack = stack[0 : len(stack) - 1]
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是top的下一位元素，i是将要入栈的元素 left := stack[len(stack) - 1]
left := stack[len(stack)-1] // 高度x宽度
// 高度x宽度 tmp := heights[mid] * (i - left - 1)
tmp := heights[mid] * (i - left - 1) if tmp > max {
if tmp > max { max = tmp
max = tmp }
} }
} stack = append(stack, i)
stack = append(stack, i) }
} return max
return max
} }
``` ```
### JavaScript: ### JavaScript: