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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

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。

给出一个数组，找到其满足摆动的子序列，最长长度。

方法一：动态规划

up[i]表示到i为止，以上升为结束的最大长度，down[i] 则相反。

class Solution {    public int wiggleMaxLength(int[] nums) {        if (nums.length <= 1) {            return nums.length;        }        int []up = new int[nums.length];        int []down = new int[nums.length];        up[0] = 1;        down[0] = 1;        for (int i = 1; i < nums.length; i++) {            if (nums[i] > nums[i - 1]) {                up[i] = down[i - 1] + 1;                down[i] = down[i - 1];            } else if (nums[i] < nums[i - 1]){                up[i] = up[i - 1];                down[i] = up[i - 1] + 1;            } else {                up[i] = up[i - 1];                down[i] = down[i - 1];            }        }        return Math.max(up[nums.length - 1], down[nums.length - 1]);    }}

方法二：找出拐点的数量即可。

class Solution {    public int wiggleMaxLength(int[] nums) {        if (nums.length <= 1) {            return nums.length;        }        boolean isUp = false;        int index = 1;        int ans = 1;// 找到第一处拐点        for (; index < nums.length; index++) {            if (nums[index] > nums[index - 1]) {                isUp = true;                ans = 2;                break;            } else if (nums[index] < nums[index - 1]){                isUp = false;                ans = 2;                break;            }        }// 找出所有拐点的数量        for (int i = index; i < nums.length; i++) {            if (nums[i] > nums[i - 1]) {                if (!isUp) {                    ans++;                    isUp = true;                }            } else if (nums[i] < nums[i - 1]) {                if (isUp) {                    ans++;                    isUp = false;                }            }        }        return ans;    }}