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Merge pull request #1267 from dmzlingyin/master
update (0739.每日温度.md): 增加未精简版本Go代码
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@ -244,6 +244,39 @@ class Solution:
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```
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Go:
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> 未精简版本
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```go
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func nextGreaterElement(nums1 []int, nums2 []int) []int {
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res := make([]int, len(nums1))
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for i := range res { res[i] = -1 }
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m := make(map[int]int, len(nums1))
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for k, v := range nums1 { m[v] = k }
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stack := []int{0}
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for i := 1; i < len(nums2); i++ {
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top := stack[len(stack)-1]
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if nums2[i] < nums2[top] {
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stack = append(stack, i)
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} else if nums2[i] == nums2[top] {
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stack = append(stack, i)
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} else {
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for len(stack) != 0 && nums2[i] > nums2[top] {
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if v, ok := m[nums2[top]]; ok {
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res[v] = nums2[i]
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}
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stack = stack[:len(stack)-1]
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if len(stack) != 0 {
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top = stack[len(stack)-1]
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}
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}
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stack = append(stack, i)
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}
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}
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return res
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}
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```
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> 精简版本
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```go
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func nextGreaterElement(nums1 []int, nums2 []int) []int {
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res := make([]int, len(nums1))
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@ -34,7 +34,7 @@
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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?
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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。
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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。
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在使用单调栈的时候首先要明确如下几点:
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@ -233,7 +233,7 @@ class Solution {
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}
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```
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Python:
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``` Python3
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```python
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class Solution:
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def dailyTemperatures(self, temperatures: List[int]) -> List[int]:
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answer = [0]*len(temperatures)
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@ -277,8 +277,36 @@ func dailyTemperatures(t []int) []int {
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}
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```
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> 单调栈法
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> 单调栈法(未精简版本)
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```go
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func dailyTemperatures(temperatures []int) []int {
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res := make([]int, len(temperatures))
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// 初始化栈顶元素为第一个下标索引0
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stack := []int{0}
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for i := 1; i < len(temperatures); i++ {
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top := stack[len(stack)-1]
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if temperatures[i] < temperatures[top] {
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stack = append(stack, i)
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} else if temperatures[i] == temperatures[top] {
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stack = append(stack, i)
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} else {
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for len(stack) != 0 && temperatures[i] > temperatures[top] {
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res[top] = i - top
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stack = stack[:len(stack)-1]
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if len(stack) != 0 {
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top = stack[len(stack)-1]
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}
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}
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stack = append(stack, i)
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}
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}
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return res
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}
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```
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> 单调栈法(精简版本)
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```go
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// 单调递减栈
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func dailyTemperatures(num []int) []int {
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