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yanglbmeyanglbme
committedJul 22, 2021
feat: add solutions to lc problem: No.0376. Wiggle Subsequence
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‎README.md

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- [最长回文子串](./solution/0000-0099/0005.Longest%20Palindromic%20Substring/README.md)
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- [最长回文子序列](.solution/0500-0599/0516.Longest%20Palindromic%20Subsequence/README.md)
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- [最长递增子序列](./solution/0300-0399/0300.Longest%20Increasing%20Subsequence/README.md)
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- [摆动序列](./solution/0300-0399/0376.Wiggle%20Subsequence/README.md)
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- [礼物的最大价值](./lcof/面试题47.%20礼物的最大价值/README.md)
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- [俄罗斯套娃信封问题](./solution/0300-0399/0354.Russian%20Doll%20Envelopes/README.md)
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- [最长公共子序列](./solution/1100-1199/1143.Longest%20Common%20Subsequence/README.md)

‎README_EN.md

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@@ -189,6 +189,7 @@ Complete solutions to [LeetCode](https://leetcode.com/problemset/all/), [LCOF](h
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- [Longest Palindromic Substring](./solution/0000-0099/0005.Longest%20Palindromic%20Substring/README_EN.md)
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- [Longest Palindromic Subsequence](.solution/0500-0599/0516.Longest%20Palindromic%20Subsequence/README_EN.md)
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- [Longest Increasing Subsequence](./solution/0300-0399/0300.Longest%20Increasing%20Subsequence/README_EN.md)
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- [Wiggle Subsequence](./solution/0300-0399/0376.Wiggle%20Subsequence/README_EN.md)
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- [Russian Doll Envelopes](./solution/0300-0399/0354.Russian%20Doll%20Envelopes/README_EN.md)
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- [Longest Common Subsequence](./solution/1100-1199/1143.Longest%20Common%20Subsequence/README_EN.md)
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‎solution/0300-0399/0376.Wiggle Subsequence/README.md

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<p><strong>进阶:</strong><br>
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你能否用&nbsp;O(<em>n</em>) 时间复杂度完成此题?</p>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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动态规划。
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设 up 表示以前 i 个元素中的某一个元素结尾的最长上升摆动序列的长度，down 表示以前 i 个元素中的某一个元素结尾的最长下降摆动序列的长度。初始 `up = 1`, `down = 1`。
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从数组下标 1 开始遍历：
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- 若 `nums[i] > nums[i - 1]`，则需要更新最长上升摆动序列的长度：`up = max(up, down + 1)`
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- 若 `nums[i] < nums[i - 1]`，则需要更新最长下降摆动序列的长度：`down = max(down, up + 1)`
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最后返回 `max(up, down)` 即可。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def wiggleMaxLength(self, nums: List[int]) -> int:
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up = down = 1
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for i in range(1, len(nums)):
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if nums[i] > nums[i - 1]:
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up = max(up, down + 1)
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elif nums[i] < nums[i - 1]:
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down = max(down, up + 1)
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return max(up, down)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int wiggleMaxLength(int[] nums) {
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int up = 1, down = 1;
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for (int i = 1; i < nums.length; ++i) {
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if (nums[i] > nums[i - 1]) {
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up = Math.max(up, down + 1);
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} else if (nums[i] < nums[i - 1]) {
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down = Math.max(down, up + 1);
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}
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}
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return Math.max(up, down);
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int wiggleMaxLength(vector<int>& nums) {
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int up = 1, down = 1;
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for (int i = 1; i < nums.size(); ++i) {
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if (nums[i] > nums[i - 1]) {
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up = max(up, down + 1);
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} else if (nums[i] < nums[i - 1]) {
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down = max(down, up + 1);
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}
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}
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return max(up, down);
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}
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};
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```
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### **Go**
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```go
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func wiggleMaxLength(nums []int) int {
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up, down := 1, 1
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for i := 1; i < len(nums); i++ {
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if nums[i] > nums[i-1] {
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up = max(up, down+1)
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} else if nums[i] < nums[i-1] {
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down = max(down, up+1)
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}
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}
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return max(up, down)
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/0300-0399/0376.Wiggle Subsequence/README_EN.md

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@@ -51,21 +51,84 @@ One is [1, 17, 10, 13, 10, 16, 8] with differences (16, -7, 3, -3, 6, -8).
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<p>&nbsp;</p>
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<p><strong>Follow up:</strong> Could you solve this in <code>O(n)</code> time?</p>
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## Solutions
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Dynamic programming.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def wiggleMaxLength(self, nums: List[int]) -> int:
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up = down = 1
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for i in range(1, len(nums)):
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if nums[i] > nums[i - 1]:
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up = max(up, down + 1)
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elif nums[i] < nums[i - 1]:
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down = max(down, up + 1)
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return max(up, down)
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```
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### **Java**
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```java
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class Solution {
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public int wiggleMaxLength(int[] nums) {
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int up = 1, down = 1;
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for (int i = 1; i < nums.length; ++i) {
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if (nums[i] > nums[i - 1]) {
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up = Math.max(up, down + 1);
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} else if (nums[i] < nums[i - 1]) {
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down = Math.max(down, up + 1);
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}
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}
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return Math.max(up, down);
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int wiggleMaxLength(vector<int>& nums) {
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int up = 1, down = 1;
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for (int i = 1; i < nums.size(); ++i) {
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if (nums[i] > nums[i - 1]) {
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up = max(up, down + 1);
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} else if (nums[i] < nums[i - 1]) {
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down = max(down, up + 1);
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}
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}
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return max(up, down);
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}
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};
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```
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### **Go**
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```go
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func wiggleMaxLength(nums []int) int {
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up, down := 1, 1
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for i := 1; i < len(nums); i++ {
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if nums[i] > nums[i-1] {
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up = max(up, down+1)
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} else if nums[i] < nums[i-1] {
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down = max(down, up+1)
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}
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}
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return max(up, down)
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/0300-0399/0376.Wiggle Subsequence/Solution.cpp

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class Solution {
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public:
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int wiggleMaxLength(vector<int>& nums) {
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int up = 1, down = 1;
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for (int i = 1; i < nums.size(); ++i) {
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if (nums[i] > nums[i - 1]) {
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up = max(up, down + 1);
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} else if (nums[i] < nums[i - 1]) {
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down = max(down, up + 1);
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}
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}
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return max(up, down);
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}
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};

‎solution/0300-0399/0376.Wiggle Subsequence/Solution.go

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func wiggleMaxLength(nums []int) int {
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up, down := 1, 1
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for i := 1; i < len(nums); i++ {
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if nums[i] > nums[i-1] {
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up = max(up, down+1)
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} else if nums[i] < nums[i-1] {
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down = max(down, up+1)
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}
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}
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return max(up, down)
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}

‎solution/0300-0399/0376.Wiggle Subsequence/Solution.java

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class Solution {
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public int wiggleMaxLength(int[] nums) {
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if (nums.length < 2) {
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return nums.length;
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}
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int up = 1;
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int down = 1;
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int up = 1, down = 1;
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for (int i = 1; i < nums.length; ++i) {
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if (nums[i] > nums[i - 1]) {
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up = down + 1;
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up = Math.max(up, down + 1);
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} else if (nums[i] < nums[i - 1]) {
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down = up + 1;
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down = Math.max(down, up + 1);
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}
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}
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return Math.max(up, down);
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}
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}
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}

‎solution/0300-0399/0376.Wiggle Subsequence/Solution.py

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class Solution:
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def wiggleMaxLength(self, nums: List[int]) -> int:
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up = down = 1
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for i in range(1, len(nums)):
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if nums[i] > nums[i - 1]:
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up = max(up, down + 1)
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elif nums[i] < nums[i - 1]:
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down = max(down, up + 1)
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return max(up, down)

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