Add solution 376

leetcode/0376.Wiggle-Subsequence/376. Wiggle Subsequence.go

@@ -0,0 +1,20 @@
+package leetcode
+
+func wiggleMaxLength(nums []int) int {
+	if len(nums) < 2 {
+		return len(nums)
+	}
+	res := 1
+	prevDiff := nums[1] - nums[0]
+	if prevDiff != 0 {
+		res = 2
+	}
+	for i := 2; i < len(nums); i++ {
+		diff := nums[i] - nums[i-1]
+		if diff > 0 && prevDiff <= 0 || diff < 0 && prevDiff >= 0 {
+			res++
+			prevDiff = diff
+		}
+	}
+	return res
+}

leetcode/0376.Wiggle-Subsequence/376. Wiggle Subsequence_test.go

@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question376 struct {
+	para376
+	ans376
+}
+
+// para 是参数
+// one 代表第一个参数
+type para376 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans376 struct {
+	one int
+}
+
+func Test_Problem376(t *testing.T) {
+
+	qs := []question376{
+
+		{
+			para376{[]int{1, 7, 4, 9, 2, 5}},
+			ans376{6},
+		},
+
+		{
+			para376{[]int{1, 17, 5, 10, 13, 15, 10, 5, 16, 8}},
+			ans376{7},
+		},
+
+		{
+			para376{[]int{1, 2, 3, 4, 5, 6, 7, 8, 9}},
+			ans376{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 376------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans376, q.para376
+		fmt.Printf("【input】:%v       【output】:%v\n", p, wiggleMaxLength(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0376.Wiggle-Subsequence/README.md

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+# [376. Wiggle Subsequence](https://leetcode.com/problems/wiggle-subsequence/)
+
+
+## 题目
+
+Given an integer array `nums`, return *the length of the longest **wiggle sequence***.
+
+A **wiggle sequence** is a sequence where the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with fewer than two elements is trivially a wiggle sequence.
+
+- For example, `[1, 7, 4, 9, 2, 5]` is a **wiggle sequence** because the differences `(6, -3, 5, -7, 3)` are alternately positive and negative.
+- In contrast, `[1, 4, 7, 2, 5]` and `[1, 7, 4, 5, 5]` are not wiggle sequences, the first because its first two differences are positive and the second because its last difference is zero.
+
+A **subsequence** is obtained by deleting some elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order.
+
+**Example 1:**
+
+```
+Input: nums = [1,7,4,9,2,5]
+Output: 6
+Explanation: The entire sequence is a wiggle sequence.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,17,5,10,13,15,10,5,16,8]
+Output: 7
+Explanation: There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,2,3,4,5,6,7,8,9]
+Output: 2
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 1000`
+- `0 <= nums[i] <= 1000`
+
+**Follow up:** Could you solve this in `O(n)` time?
+
+## 题目大意
+
+如果连续数字之间的差严格地在正数和负数之间交替，则数字序列称为摆动序列。第一个差（如果存在的话）可能是正数或负数。少于两个元素的序列也是摆动序列。例如， [1,7,4,9,2,5] 是一个摆动序列，因为差值 (6,-3,5,-7,3) 是正负交替出现的。相反, [1,4,7,2,5] 和 [1,7,4,5,5] 不是摆动序列，第一个序列是因为它的前两个差值都是正数，第二个序列是因为它的最后一个差值为零。给定一个整数序列，返回作为摆动序列的最长子序列的长度。 通过从原始序列中删除一些（也可以不删除）元素来获得子序列，剩下的元素保持其原始顺序。
+
+## 解题思路
+
+- 题目要求找到摆动序列最长的子序列。本题可以用贪心的思路，记录当前序列的上升和下降的趋势。扫描数组过程中，每扫描一个元素都判断是“峰”还是“谷”，根据前一个是“峰”还是“谷”做出对应的决定。利用贪心的思想找到最长的摆动子序列。
+
+## 代码
+
+```go
+package leetcode
+
+func wiggleMaxLength(nums []int) int {
+	if len(nums) < 2 {
+		return len(nums)
+	}
+	res := 1
+	prevDiff := nums[1] - nums[0]
+	if prevDiff != 0 {
+		res = 2
+	}
+	for i := 2; i < len(nums); i++ {
+		diff := nums[i] - nums[i-1]
+		if diff > 0 && prevDiff <= 0 || diff < 0 && prevDiff >= 0 {
+			res++
+			prevDiff = diff
+		}
+	}
+	return res
+}
+```