A sequence of numbers is called a wiggle sequence if 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.
Given a sequence of integers, return the length of the longest subsequence that is a wiggle sequence. A subsequence is obtained by deleting some number of elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order.
Example 1:
Input: [1,7,4,9,2,5]
Output: 6
Explanation: The entire sequence is a wiggle sequence.
Example 2:
Input: [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: [1,2,3,4,5,6,7,8,9]
Output: 2
Follow up:
Can you do it in O(n) time?
摆动序列。
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。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/wiggle-subsequence
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
思路是动态规划。这道题的思路很像300题最长上升子序列，我们创建两个数组up和down，表示以nums[i]为结尾的，最后是上升/下降的最长子序列的长度。
如果nums[i] > nums[i - 1]，说明最后是上升的，最长子序列的长度是down[i - 1] + 1，因为最后是上升的话，再往前应该是下降的
如果nums[i] < nums[i - 1]，说明最后是下降的，最长子序列的长度是up[i - 1] + 1，因为最后是下降的话，再往前应该是上升的
如果nums[i] == nums[i - 1]，说明已经破坏了上升/下降的规则，则最长子序列的长度跟前一个位置相同
时间O(n)
空间O(n)
Java实现
1 class Solution { 2 public int wiggleMaxLength(int[] nums) { 3 // corner case 4 if (nums.length <= 1) { 5 return nums.length; 6 } 7 8 // normal case 9 int len = nums.length; 10 int[] down = new int[len]; 11 int[] up = new int[len]; 12 down[0] = 1; 13 up[0] = 1; 14 for (int i = 1; i < len; i++) { 15 if (nums[i] > nums[i - 1]) { 16 down[i] = down[i - 1]; 17 up[i] = down[i - 1] + 1; 18 } else if (nums[i] < nums[i - 1]) { 19 down[i] = up[i - 1] + 1; 20 up[i] = up[i - 1]; 21 } else { 22 down[i] = down[i - 1]; 23 up[i] = up[i - 1]; 24 } 25 } 26 return Math.max(down[len - 1], up[len - 1]); 27 } 28 }
另外提供一个不使用额外空间的做法。
1 class Solution { 2 public int wiggleMaxLength(int[] nums) { 3 // corner case 4 if (nums.length <= 1) { 5 return nums.length; 6 } 7 8 // normal case 9 int up = 1; 10 int down = 1; 11 for (int i = 1; i < nums.length; i++) { 12 if (nums[i] > nums[i - 1]) { 13 up = down + 1; 14 } else if (nums[i] < nums[i - 1]) { 15 down = up + 1; 16 } 17 } 18 return Math.max(up, down); 19 } 20 }