mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2025-07-06 17:44:10 +08:00
fix 0978: update a clearer solution
This commit is contained in:
novahe
@ -1,14 +1,14 @@
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package leetcode
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// 解法一 模拟法
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func maxTurbulenceSize(A []int) int {
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func maxTurbulenceSize(arr []int) int {
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inc, dec := 1, 1
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maxLen := min(1, len(A))
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for i := 1; i < len(A); i++ {
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if A[i-1] < A[i] {
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maxLen := min(1, len(arr))
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for i := 1; i < len(arr); i++ {
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if arr[i-1] < arr[i] {
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inc = dec + 1
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dec = 1
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} else if A[i-1] > A[i] {
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} else if arr[i-1] > arr[i] {
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dec = inc + 1
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inc = 1
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} else {
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@ -35,23 +35,21 @@ func min(a int, b int) int {
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}
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// 解法二 滑动窗口
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func maxTurbulenceSize1(A []int) int {
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if len(A) == 1 {
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return 1
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func maxTurbulenceSize1(arr []int) int {
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var maxLength int
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if len(arr) == 2 && arr[0] != arr[1] {
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maxLength = 2
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} else {
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maxLength = 1
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}
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// flag > 0 代表下一个数要大于前一个数，flag < 0 代表下一个数要小于前一个数
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res, left, right, flag, lastNum := 0, 0, 0, A[1]-A[0], A[0]
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for left < len(A) {
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if right < len(A)-1 && ((A[right+1] > lastNum && flag > 0) || (A[right+1] < lastNum && flag < 0) || (right == left)) {
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right++
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flag = lastNum - A[right]
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lastNum = A[right]
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} else {
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if flag != 0 {
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res = max(res, right-left+1)
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}
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left++
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left := 0
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for right := 2; right < len(arr); right++ {
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if arr[right] == arr[right-1] {
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left = right
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} else if (arr[right]-arr[right-1])^(arr[right-1]-arr[right-2]) >= 0 {
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left = right - 1
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}
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maxLength = max(maxLength, right-left+1)
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}
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return max(res, 1)
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return maxLength
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}
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@ -1,58 +1,58 @@
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# [978. Longest Turbulent Subarray](https://leetcode.com/problems/longest-turbulent-subarray/)
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## 题目
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A subarray `A[i], A[i+1], ..., A[j]` of `A` is said to be *turbulent* if and only if:
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- For `i <= k < j`, `A[k] > A[k+1]` when `k` is odd, and `A[k] < A[k+1]` when `k` is even;
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- **OR**, for `i <= k < j`, `A[k] > A[k+1]` when `k` is even, and `A[k] < A[k+1]` when `k` is odd.
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That is, the subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray.
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Return the **length** of a maximum size turbulent subarray of A.
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**Example 1:**
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Input: [9,4,2,10,7,8,8,1,9]
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Output: 5
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Explanation: (A[1] > A[2] < A[3] > A[4] < A[5])
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**Example 2:**
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Input: [4,8,12,16]
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Output: 2
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**Example 3:**
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Input: [100]
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Output: 1
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**Note:**
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1. `1 <= A.length <= 40000`
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2. `0 <= A[i] <= 10^9`
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## 题目大意
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当 A 的子数组 A[i], A[i+1], ..., A[j] 满足下列条件时，我们称其为湍流子数组：
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若 i <= k < j，当 k 为奇数时， A[k] > A[k+1]，且当 k 为偶数时，A[k] < A[k+1]；
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或 若 i <= k < j，当 k 为偶数时，A[k] > A[k+1] ，且当 k 为奇数时， A[k] < A[k+1]。
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也就是说，如果比较符号在子数组中的每个相邻元素对之间翻转，则该子数组是湍流子数组。
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返回 A 的最大湍流子数组的长度。
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提示：
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- 1 <= A.length <= 40000
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- 0 <= A[i] <= 10^9
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## 解题思路
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- 给出一个数组，要求找出“摆动数组”的最大长度。所谓“摆动数组”的意思是，元素一大一小间隔的。
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- 这一题可以用滑动窗口来解答。用一个变量记住下次出现的元素需要大于还是需要小于前一个元素。也可以用模拟的方法，用两个变量分别记录上升和下降数字的长度。一旦元素相等了，上升和下降数字长度都置为 1，其他时候按照上升和下降的关系增加队列长度即可，最后输出动态维护的最长长度。
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# [978. Longest Turbulent Subarray](https://leetcode.com/problems/longest-turbulent-subarray/)
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## 题目
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A subarray `A[i], A[i+1], ..., A[j]` of `A` is said to be *turbulent* if and only if:
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- For `i <= k < j`, `A[k] > A[k+1]` when `k` is odd, and `A[k] < A[k+1]` when `k` is even;
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- **OR**, for `i <= k < j`, `A[k] > A[k+1]` when `k` is even, and `A[k] < A[k+1]` when `k` is odd.
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That is, the subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray.
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Return the **length** of a maximum size turbulent subarray of A.
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**Example 1:**
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Input: [9,4,2,10,7,8,8,1,9]
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Output: 5
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Explanation: (A[1] > A[2] < A[3] > A[4] < A[5])
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**Example 2:**
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Input: [4,8,12,16]
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Output: 2
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**Example 3:**
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Input: [100]
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Output: 1
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**Note:**
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1. `1 <= A.length <= 40000`
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2. `0 <= A[i] <= 10^9`
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## 题目大意
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当 A 的子数组 A[i], A[i+1], ..., A[j] 满足下列条件时，我们称其为湍流子数组：
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若 i <= k < j，当 k 为奇数时， A[k] > A[k+1]，且当 k 为偶数时，A[k] < A[k+1]；
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或 若 i <= k < j，当 k 为偶数时，A[k] > A[k+1] ，且当 k 为奇数时， A[k] < A[k+1]。
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也就是说，如果比较符号在子数组中的每个相邻元素对之间翻转，则该子数组是湍流子数组。
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返回 A 的最大湍流子数组的长度。
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提示：
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- 1 <= A.length <= 40000
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- 0 <= A[i] <= 10^9
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## 解题思路
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- 给出一个数组，要求找出“摆动数组”的最大长度。所谓“摆动数组”的意思是，元素一大一小间隔的。
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- 这一题可以用滑动窗口来解答。用相邻元素差的乘积大于零（a ^ b >= 0 说明a b乘积大于零）来判断是否是湍流， 如果是，那么扩大窗口。否则窗口缩小为0，开始新的一个窗口。
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@ -55,8 +55,7 @@ Return the **length** of a maximum size turbulent subarray of A.
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- 给出一个数组，要求找出“摆动数组”的最大长度。所谓“摆动数组”的意思是，元素一大一小间隔的。
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- 这一题可以用滑动窗口来解答。用一个变量记住下次出现的元素需要大于还是需要小于前一个元素。也可以用模拟的方法，用两个变量分别记录上升和下降数字的长度。一旦元素相等了，上升和下降数字长度都置为 1，其他时候按照上升和下降的关系增加队列长度即可，最后输出动态维护的最长长度。
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- 这一题可以用滑动窗口来解答。用相邻元素差的乘积大于零（a ^ b >= 0 说明a b乘积大于零）来判断是否是湍流， 如果是，那么扩大窗口。否则窗口缩小为0，开始新的一个窗口。
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## 代码
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@ -65,14 +64,14 @@ Return the **length** of a maximum size turbulent subarray of A.
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package leetcode
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// 解法一 模拟法
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func maxTurbulenceSize(A []int) int {
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func maxTurbulenceSize(arr []int) int {
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inc, dec := 1, 1
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maxLen := min(1, len(A))
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for i := 1; i < len(A); i++ {
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if A[i-1] < A[i] {
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maxLen := min(1, len(arr))
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for i := 1; i < len(arr); i++ {
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if arr[i-1] < arr[i] {
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inc = dec + 1
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dec = 1
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} else if A[i-1] > A[i] {
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} else if arr[i-1] > arr[i] {
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dec = inc + 1
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inc = 1
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} else {
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@ -85,25 +84,23 @@ func maxTurbulenceSize(A []int) int {
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}
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// 解法二 滑动窗口
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func maxTurbulenceSize1(A []int) int {
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if len(A) == 1 {
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return 1
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func maxTurbulenceSize1(arr []int) int {
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var maxLength int
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if len(arr) == 2 && arr[0] != arr[1] {
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maxLength = 2
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} else {
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maxLength = 1
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}
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// flag > 0 代表下一个数要大于前一个数，flag < 0 代表下一个数要小于前一个数
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res, left, right, flag, lastNum := 0, 0, 0, A[1]-A[0], A[0]
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for left < len(A) {
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if right < len(A)-1 && ((A[right+1] > lastNum && flag > 0) || (A[right+1] < lastNum && flag < 0) || (right == left)) {
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right++
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flag = lastNum - A[right]
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lastNum = A[right]
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} else {
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if flag != 0 {
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res = max(res, right-left+1)
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}
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left++
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left := 0
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for right := 2; right < len(arr); right++ {
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if arr[right] == arr[right-1] {
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left = right
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} else if (arr[right]-arr[right-1])^(arr[right-1]-arr[right-2]) >= 0 {
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left = right - 1
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}
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maxLength = max(maxLength, right-left+1)
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}
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return max(res, 1)
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return maxLength
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}
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```
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