mirror of
https://github.com/youngyangyang04/leetcode-master.git
synced 2025-07-08 16:54:50 +08:00
Merge pull request #1267 from dmzlingyin/master
update (0739.每日温度.md): 增加未精简版本Go代码
This commit is contained in:
@ -244,6 +244,39 @@ class Solution:
|
|||||||
```
|
```
|
||||||
|
|
||||||
Go:
|
Go:
|
||||||
|
|
||||||
|
> 未精简版本
|
||||||
|
```go
|
||||||
|
func nextGreaterElement(nums1 []int, nums2 []int) []int {
|
||||||
|
res := make([]int, len(nums1))
|
||||||
|
for i := range res { res[i] = -1 }
|
||||||
|
m := make(map[int]int, len(nums1))
|
||||||
|
for k, v := range nums1 { m[v] = k }
|
||||||
|
|
||||||
|
stack := []int{0}
|
||||||
|
for i := 1; i < len(nums2); i++ {
|
||||||
|
top := stack[len(stack)-1]
|
||||||
|
if nums2[i] < nums2[top] {
|
||||||
|
stack = append(stack, i)
|
||||||
|
} else if nums2[i] == nums2[top] {
|
||||||
|
stack = append(stack, i)
|
||||||
|
} else {
|
||||||
|
for len(stack) != 0 && nums2[i] > nums2[top] {
|
||||||
|
if v, ok := m[nums2[top]]; ok {
|
||||||
|
res[v] = nums2[i]
|
||||||
|
}
|
||||||
|
stack = stack[:len(stack)-1]
|
||||||
|
if len(stack) != 0 {
|
||||||
|
top = stack[len(stack)-1]
|
||||||
|
}
|
||||||
|
}
|
||||||
|
stack = append(stack, i)
|
||||||
|
}
|
||||||
|
}
|
||||||
|
return res
|
||||||
|
}
|
||||||
|
```
|
||||||
|
> 精简版本
|
||||||
```go
|
```go
|
||||||
func nextGreaterElement(nums1 []int, nums2 []int) []int {
|
func nextGreaterElement(nums1 []int, nums2 []int) []int {
|
||||||
res := make([]int, len(nums1))
|
res := make([]int, len(nums1))
|
||||||
|
@ -34,7 +34,7 @@
|
|||||||
|
|
||||||
那么单调栈的原理是什么呢?为什么时间复杂度是O(n)就可以找到每一个元素的右边第一个比它大的元素位置呢?
|
那么单调栈的原理是什么呢?为什么时间复杂度是O(n)就可以找到每一个元素的右边第一个比它大的元素位置呢?
|
||||||
|
|
||||||
单调栈的本质是空间换时间,因为在遍历的过程中需要用一个栈来记录右边第一个比当前元素的元素,优点是只需要遍历一次。
|
单调栈的本质是空间换时间,因为在遍历的过程中需要用一个栈来记录右边第一个比当前元素大的元素,优点是只需要遍历一次。
|
||||||
|
|
||||||
|
|
||||||
在使用单调栈的时候首先要明确如下几点:
|
在使用单调栈的时候首先要明确如下几点:
|
||||||
@ -233,7 +233,7 @@ class Solution {
|
|||||||
}
|
}
|
||||||
```
|
```
|
||||||
Python:
|
Python:
|
||||||
``` Python3
|
```python
|
||||||
class Solution:
|
class Solution:
|
||||||
def dailyTemperatures(self, temperatures: List[int]) -> List[int]:
|
def dailyTemperatures(self, temperatures: List[int]) -> List[int]:
|
||||||
answer = [0]*len(temperatures)
|
answer = [0]*len(temperatures)
|
||||||
@ -277,8 +277,36 @@ func dailyTemperatures(t []int) []int {
|
|||||||
}
|
}
|
||||||
```
|
```
|
||||||
|
|
||||||
> 单调栈法
|
> 单调栈法(未精简版本)
|
||||||
|
|
||||||
|
```go
|
||||||
|
func dailyTemperatures(temperatures []int) []int {
|
||||||
|
res := make([]int, len(temperatures))
|
||||||
|
// 初始化栈顶元素为第一个下标索引0
|
||||||
|
stack := []int{0}
|
||||||
|
|
||||||
|
for i := 1; i < len(temperatures); i++ {
|
||||||
|
top := stack[len(stack)-1]
|
||||||
|
if temperatures[i] < temperatures[top] {
|
||||||
|
stack = append(stack, i)
|
||||||
|
} else if temperatures[i] == temperatures[top] {
|
||||||
|
stack = append(stack, i)
|
||||||
|
} else {
|
||||||
|
for len(stack) != 0 && temperatures[i] > temperatures[top] {
|
||||||
|
res[top] = i - top
|
||||||
|
stack = stack[:len(stack)-1]
|
||||||
|
if len(stack) != 0 {
|
||||||
|
top = stack[len(stack)-1]
|
||||||
|
}
|
||||||
|
}
|
||||||
|
stack = append(stack, i)
|
||||||
|
}
|
||||||
|
}
|
||||||
|
return res
|
||||||
|
}
|
||||||
|
```
|
||||||
|
|
||||||
|
> 单调栈法(精简版本)
|
||||||
```go
|
```go
|
||||||
// 单调递减栈
|
// 单调递减栈
|
||||||
func dailyTemperatures(num []int) []int {
|
func dailyTemperatures(num []int) []int {
|
||||||
|
Reference in New Issue
Block a user